Can you check if this proof is correct? Im interested in shorter/alternative proofs too. Thank you.
Let $x_0,\ldots,x_k\in\Bbb K^n$ be non-zero. Show that $$A:=\left\{x\in \Bbb K^n:\prod_{j=0}^k(x\mid x_j)\neq0\right\}$$ is open and dense in $\Bbb K^n$.
The inner product is a continuous function and the set $A$ is the preimage of $\Bbb K\setminus\{0\}$, which is open, hence $A$ is open.
Now remember that
$$(x\mid\alpha y)=\bar\alpha(x\mid y),\quad\forall \alpha\in\Bbb K,\forall x,y\in\Bbb K^n$$
then
$$(x\mid y)=0\implies(x\mid\alpha y)=0\tag{1}$$
for all $\alpha\in\Bbb K$. Let an arbitrary $y\in\Bbb K^n$ and the open ball $\Bbb B (y,\varepsilon)$ for some arbitrary $\varepsilon>0$. If $A$ is dense in $\Bbb K^n$ then exists some $z\in\Bbb B(y,\varepsilon)$ such that $z\in A$.
Part one: if $y\in A$ then for some suitable $\alpha>0$ we have that $\alpha y\in\Bbb B(y,\varepsilon)$, $y\neq 0$ and $\prod (\alpha y\mid x_j)\neq 0$. Proof: first note that if $y=0$ then $(y\mid z)=0$ for all $z\in\Bbb K^n$, what would imply that $y\neq A$, hence if $y\in A$ then $y\neq 0$. Now observe that
$$d(\alpha y,y)=\sqrt{(\alpha y-y\mid\alpha y-y)}=|\alpha-1|\|y\|$$
hence choosing $\alpha\in(1-\varepsilon/\|y\|,1+\varepsilon/\|y\|)$ we have that
$$d(\alpha y,y)<\varepsilon$$
what imply that $\alpha y\in\Bbb B(y,\varepsilon)$. And
$$\prod_{j=0}^n(\alpha y\mid x_j)=\alpha^{n+1}\prod_{j=0}^n(y\mid x_j)\neq 0$$
$\Box$
Part two: suppose that $y\notin A$ and $y\neq 0$, hence exists one or more $x_j$ such that $(y\mid x_j)=0$. Then I claim that exists some $z\in A$ such that $z\in\Bbb B(y,\varepsilon)$. Proof: choose some $z\in\Bbb B(y,\varepsilon)$, that is
$$d(z,y)=\sqrt{(z-y\mid z-y)}=\sqrt{\|z\|^2+\|y\|^2-2\Re(z\mid y)}<\varepsilon\tag{2}$$
If $(z\mid y)\neq 0$ we are done, then suppose that $(z\mid y)=0$, but observe that
$$((\beta-1) z+\beta y\mid y)=((\beta-1) z\mid y)+\beta(y\mid y)=0+\beta \|y\|^2\neq 0$$
for $\beta\neq 0$, and we have that
$$\begin{align}d((\beta-1) z+\beta y,y)&=\sqrt{((\beta-1) z+\beta y-y\mid(\beta-1) z+\beta y-y)}\\&=|\beta-1|\sqrt{\|z\|^2+\|y\|^2-2\Re (z\mid y)}\end{align}\tag{3}$$
Then comparing (2) and (3) is clear that exists some $\beta>1$ such that $d(z,y)>d((\beta-1)z+\beta y,y) $, thus $(\beta-1)z+\beta y\in\Bbb B(y,\varepsilon)$ and $(\beta-1)z+\beta y\in A$, hence $A$ is dense in $\Bbb K^n$.$\Box$
Part three: finally it remain to show that for any $\epsilon$-neighborhood of the zero vector, i.e. for some $\Bbb B_\varepsilon:=\Bbb B(0,\varepsilon)$, exists some $z\in A$ such that $z\in\Bbb B_\varepsilon$.
I claim that $A\neq \emptyset$ and that for any $z\in A$ exists some $z^*\in A$ such that $z^*\in\Bbb B_\varepsilon$. Proof: define $z:=\sum_{j=0}^k x_j$, then its easy to check that
$$(z\mid x_j)\neq 0,\quad\forall x_j\in B$$
provided that $\|x_j\|\neq 0$, what is an assumption to the exercise, hence $A\neq\emptyset$. Now observe that for any $z\in A$ we have that $\|z\|\neq 0$ (proved in the part one), then define $z^*:=\delta z/\|z\|$ such that $0<\delta<\varepsilon$, hence $\|z^*\|=\delta$, what implies that $z^*\in\Bbb B_\varepsilon$.$\Box$