1. Proof that $K$ is a compact. Consider arbitrary infinite set $\{y_k:k\in\mathbb{N}\}\subset K$, then it has the form $y_k=x_{n_k}/n_k$ for some sequence $\{n_k:k\in\mathbb{N}\}$. Since $\{y_k:k\in\mathbb{N}\}$ is infinite, then does $\{n_k:k\in\mathbb{N}\}$, hence there exist subsequence $\{n_{k_l}:l\in\mathbb{N}\}$ tending to infinity. Now note that $ \lim\limits_{l\to\infty}\Vert y_{k_l}\Vert= \lim\limits_{l\to\infty}\Vert n_{k_l}^{-1}x_{n_{k_l}}\Vert\leq \lim\limits_{l\to\infty}n_{k_l}^{-1}\left(\limsup\limits_{l\to\infty}\Vert x_{n_{k_l}}\Vert\right)=0 $ This means that $\{y_k:k\in\mathbb{N}\}$ have limit convergent subsequence $\{y_{k_l}:l\in\mathbb{N}\}$. Since $\{y_k:k\in\mathbb{N}\}\subset K$, then $K$ is a compact.
2. Proof that $co(K)$ is bounded. Consider arbitrary $x\in co(K)$, then $ x=\sum\limits_{k=1}^{m-1}\lambda_k n_k^{-1}x_{n_k}+\left(1-\sum\limits_{k=1}^{m-1}\lambda_{k}\right)0 $ for some real numbers $\{\lambda_k\in[0,1]:k\in\{1,\ldots,m\}\}$ such that $\sum_{k=1}^m\lambda_k\leq 1$ and some indexes $\{n_k\in\mathbb{N}:k\in\{1,\ldots,m-1\}\}$. In this case $ \Vert x\Vert=\left\Vert\sum\limits_{k=1}^{m-1}\lambda_k x_{n_k}\right\Vert\leq \sum\limits_{k=1}^{m-1}|\lambda_k| \Vert n_k^{-1}x_{n_k}\Vert\leq \sum\limits_{k=1}^{m-1}|\lambda_k| n_k^{-1}\leq \sum\limits_{k=1}^{m-1} \lambda_k\leq 1 $ Since $x\in co(K)$ is arbitrary, then $co(K)$ is bounded.
3. Proof that $co(K)$ is not closed. Note that the series $ \sum\limits_{n=1}^\infty\Vert 2^{-n} n^{-1} x_n\Vert\leq\sum\limits_{n=1}^\infty 2^{-n}=1 $ is convergent and we are working in a Hilbert space, hence we have well defined vector $ \hat{x}=\sum\limits_{n=1}^\infty 2^{-n} n^{-1} x_n $ From definition of $co(K)$ we see that each element of $co(K)$ is finite linear combination of elemnts of $K$, hence $\hat{x}\notin co(K)$. But $x\in\overline{co(K)}$. Indeed for each $\varepsilon>0$ consider natural number $m=\lceil\log_2 \varepsilon^{-1}\rceil+1$ and the vector $ x_\varepsilon=\sum\limits_{n=1}^{m-1} 2^{-n}n^{-1}x_n+\left(1-\sum\limits_{n=1}^{m-1} 2^{-n}\right) 0\in co(K) $ Now we see that $ \Vert \hat{x}-x_\varepsilon\Vert=\left\Vert\sum\limits_{n=m+1}^\infty 2^{-n}n^{-1}x_n\right\Vert\leq \sum\limits_{n=m+1}^\infty 2^{-n}n^{-1}\Vert x_n\Vert\leq \sum\limits_{n=m+1}^\infty 2^{-n}=2^{-m}<\varepsilon $ Since $\varepsilon>0$ is arbitrary then $\hat{x}\in\overline{co(K)}$, though $\hat{x}\notin co(K)$. Hence $co(K)$ is not closed.
4. Description of elements of $\overline{co(K)}$. Since $x\in\overline{co(K)}\subset\overline{\mathrm{span}\{x_n:n\in\mathbb{N}\}}$ and the system $\{x_n:n\in\mathbb{N}\}$ is orthonormal, then $x=\sum_{n=1}^\infty\mu_n x_n$ for some real numbers $\{\mu_n\in\mathbb{R}:n\in\mathbb{N}\}$. But for convenience we will consider the following representation $ x=\sum_{n=1}^\infty\lambda_n n^{-1} x_n $ for some real numbers $\{\lambda_n\in\mathbb{R}:n\in\mathbb{N}\}$. We claim that for all $n\in\mathbb{N}$ we have $\lambda_n\in[0,1]$ and $\sum_{n=1}^\infty\lambda_n\in(0,1)$. Fix arbitrary $\varepsilon>0$. Since $x\in\overline{co(K)}$, then there exist real numbers $\{\lambda_n^{\varepsilon}\in[0,1]:n\in\{1,\ldots,m^{\varepsilon}\}\}$ such that $ \left\Vert x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n\right\Vert\leq\varepsilon,\qquad \sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}\in[0,1] $ Then from orthogonality for all $k\in\mathbb{N}$ we get $ |\lambda_k-\lambda_k^{\varepsilon}|= \left|\langle x, k x_k\rangle-\langle \sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, k x_k\rangle\right|= k\left|\langle x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, x_k\rangle\right|\leq $ $ k\left\Vert x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n \right\Vert\Vert x_k\Vert\leq k\varepsilon $ Using the fact $\lambda_k^{\varepsilon}\in[0,1]$ we see that $ -k\varepsilon<\lambda_k^{\varepsilon}-k\varepsilon<\lambda_k<\lambda_k^{\varepsilon}+k\varepsilon<1+k\varepsilon $ Since $\varepsilon>0$ is arbitrary we conclude $\lambda_k\in[0,1]$. Since $k\in\mathbb{N}$ is also arbitrary we have $\{\lambda_n:n\in\mathbb{N}\}\subset[0,1]$.
Similarly, for all $k\in\mathbb{N}$ from orthogonality we get $ \left|\sum\limits_{n=1}^k\lambda_k - \sum\limits_{n=1}^k\lambda_n^{\varepsilon}\right|= \left|\langle x, \sum\limits_{l=1}^k l x_l\rangle-\langle \sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, \sum\limits_{l=1}^k l x_l\rangle\right|= \left|\langle x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, \sum\limits_{l=1}^k l x_l\rangle\right|\leq $ $ \left\Vert x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n\right\Vert \left\Vert \sum\limits_{l=1}^{k} l x_l\right\Vert\leq \varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2} $ Using the fact that $0\leq\sum_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}\leq 1$ we see that $ -\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2}<\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}-\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2}<\sum\limits_{n=1}^k\lambda_k<\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}+\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2}<1+\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2} $ Since $\varepsilon>0$ is arbitrary we conclude $\sum_{n=1}^k\lambda_k\in[0,1]$. Since $k\in\mathbb{N}$ is also arbitrary we have $\sum_{n=1}^\infty\lambda_k\in[0,1]$. Thus we proved that each element $x\in \overline{co(K)}$ has the form $ x=\sum_{n=1}^\infty\lambda_n n^{-1} x_n $ for some real numbers $\{\lambda_n\in[0,1]:n\in\mathbb{N}\}$ satisfying $\sum_{n=1}^\infty\lambda_k\in[0,1]$.
5. Description of extreme points of $\overline{co(K)}$. Fix some $x\in \overline{co(K)}$ and consider its representation given above.
Case 1. Consider case when $\sum_{n=1}^\infty\lambda_n\in(0,1)$, then there exist some $k\in\mathbb{N}$ such that $\lambda_k\in(0,1)$. Define $ \alpha=\min\left(\frac{1}{2},\left(\sum\limits_{n=1}^\infty\lambda_n\right)^{-1}-1\right) \qquad \delta_n= \begin{cases} \min(\lambda_n,1-\lambda_n) &\text{ if }\quad \lambda_n\in(0,1)\\ 0 &\text{ if }\quad \lambda_n\in\{0,1\} \end{cases} $ $ x^{(1)}=\sum\limits_{n=1}^\infty\lambda_n^{(1)} n^{-1} x_n\quad\text{ where }\quad\lambda_n^{(1)}=\lambda_n+\alpha\delta_n $ $ x^{(2)}=\sum\limits_{n=1}^\infty\lambda_n^{(2)} n^{-1} x_n\quad\text{ where }\quad\lambda_n^{(2)}=\lambda_n-\alpha\delta_n $ Now we check that $x^{(1)}$ and $x^{(2)}$ are in $\overline{co(K)}$. Lets go $ \lambda_n^{(1)}=\lambda_n+\alpha\delta_n>\lambda_n\geq 0 $ $ \lambda_n^{(1)}=\lambda_n+\alpha\delta_n\leq\lambda_n+\frac{1}{2}\delta_n\leq\lambda_n+\delta_n\leq\lambda_n+1-\lambda_n=1 $ $ \lambda_n^{(2)}=\lambda_n-\alpha\delta_n\geq\lambda_n-\frac{1}{2}\delta_n\geq\lambda_n-\frac{1}{2}\lambda_n=\frac{1}{2}\lambda_n\geq 0 $ $ \lambda_n^{(2)}=\lambda_n-\alpha\delta_n\leq\lambda_n\leq 1 $ $ \sum\limits_{n=1}^\infty\lambda_n^{(1)}=\sum\limits_{n=1}^\infty(\lambda_n+\alpha\delta_n)= \sum\limits_{n=1}^\infty\lambda_n+\alpha\sum\limits_{n=1}^\infty\delta_n\leq \sum\limits_{n=1}^\infty\lambda_n+\alpha\sum\limits_{n=1}^\infty\lambda_n= (\alpha+1)\sum\limits_{n=1}^\infty\lambda\leq 1 $ $ \sum\limits_{n=1}^\infty\lambda_n^{(2)}=\sum\limits_{n=1}^\infty(\lambda_n-\alpha\delta_n)\leq \sum\limits_{n=1}^\infty\lambda_n\leq 1 $ Since all necessary conditions are satisfied $x^{(1)}$, $x^{(2)}\in \overline{co(K)}$. This elements are distinct. Indeed, from orthogonality we have $ \Vert x^{(1)}-x^{(2)}\Vert^2=\left\Vert \sum\limits_{n=1}^\infty 2\alpha\delta_n x_n\right\Vert^2= 4\alpha^2\delta_n^2\sum\limits_{n=1}^\infty\Vert x_n\Vert^2> 4\alpha^2\delta_k^2\Vert x_k\Vert^2>0 $ hence $x^{(1)}\neq x^{(2)}$. And finally we see that $ x=\sum\limits_{n=1}^\infty \lambda_n n^{-1} x_n= \frac{1}{2}\sum\limits_{n=1}^\infty \lambda_n^{(1)} n^{-1} x_n+ \frac{1}{2}\sum\limits_{n=1}^\infty \lambda_n^{(2)} n^{-1} x_n=\frac{x^{(1)}+x^{(2)}}{2}. $ This means that $x$ can be middle of some non-trivial segment $[x^{(2)},x^{(2)}]$ with ends in $\overline{co(K)}$, so $x$ is not an extreme point.
Case 2. Consider case, when $\sum_{n=1}^\infty\lambda_n=0$, i.e. $x=0$. Assume we are given representation $x=t x^{(1)}+(1-t)x^{(2)}$ with $t\in(0,1)$ and $ x^{(1)}=\sum\limits_{n=1}^\infty\lambda_n^{(1)} n^{-1} x_n\qquad x^{(2)}=\sum\limits_{n=1}^\infty\lambda_n^{(2)} n^{-1} x_n $ where $0\leq\sum_{n=1}^\infty\lambda_n^{(1)}\leq 1$ and $0\leq\sum_{n=1}^\infty\lambda_n^{(2)}\leq 1$. Then from orthogonality for all $k\in\mathbb{N}$ we obtain $ t\lambda_r^{(1)}+(1-t)\lambda_r^{(2)}=t\langle x^{(1)},x_r\rangle+(1-t)\langle x^{(1)},x_r\rangle=\langle tx^{(1)}+(1-t)x^{(2)},x_r\rangle=\langle x, x_r\rangle=0 $ Because $t\in(0,1)$ and $\lambda_n^{(1)},\lambda_n^{(2)}\geq 0$ for all $n\in\mathbb{N}$ we conclude that the last equality is possible iff $\lambda_r^{(1)},\lambda_r^{(2)}=0$ for all $r\in\mathbb{N}$. As the consequence $x^{(1)}=x^{(2)}=0$. This means that $x=0$ is an extreme point.
Case 3. Now the third case, when $\sum_{n=1}^\infty\lambda_n=1$, i. e. $\lambda_n=\delta_{n,k}$ for some $k\in\mathbb{N}$. In this case $x=\sum_{n=1}^\infty\lambda_n n^{-1} x_n=k^{-1} x_k$. Again assume we are given representation $x=t x^{(1)}+(1-t)x^{(2)}$ with $t\in(0,1)$ and $ x^{(1)}=\sum\limits_{n=1}^\infty\lambda_n^{(1)} n^{-1} x_n\qquad x^{(2)}=\sum\limits_{n=1}^\infty\lambda_n^{(2)} n^{-1} x_n $ where $0\leq\sum_{n=1}^\infty\lambda_n^{(1)}\leq 1$ and $0\leq\sum_{n=1}^\infty\lambda_n^{(2)}\leq 1$. Then from orthogonality for all $k\in\mathbb{N}$ we obtain $ t\lambda_r^{(1)}+(1-t)\lambda_r^{(2)}=t\langle x^{(1)},x_r\rangle+(1-t)\langle x^{(1)},x_r\rangle=\langle tx^{(1)}+(1-t)x^{(2)},x_r\rangle=\langle x, x_r\rangle=\delta_{r,k} $ Because $t\in(0,1)$ and $\lambda_n^{(1)},\lambda_n^{(2)}\geq 0$ for all $n\in\mathbb{N}$ we conclude that the last equality is possible iff $\lambda_r^{(1)}=\lambda_r^{(2)}=\delta_{r,k}$ and as the consequence $x^{(1)}=x^{(2)}=k^{-1}x_k$. This means that $x=x_k$ is an extreme point.
Whew, all cases are considered and we conclude that extreme points of $\overline{co(K)}$ are $0$ and $\{n^{-1} x_n:n\in\mathbb{N}\}$, i.e. a set $K$.