This is counterexample for the original question.
Consider $K_1=\{0\}$ and $\eta_j=j^{-2}$, then
$$
\bigcup\limits_{j=1}^{\infty} K_j=
\mathrm{Ball}\left(0,\sum\limits_{j=1}^\infty j^{-2}\right)=
\mathrm{Ball}\left(0,\frac{\pi^2}{6}\right)
$$
If $E$ is infinite dimensional any ball is not relatively compact.
This is answer to the edited question.
Fix $\varepsilon>0$. We know that $K_j$ is a compact for all $j\in\mathbb{N}$, hence for all $j\in\mathbb{N}$ there exist $\{x_{j,l}:l=1,\ldots,N_j\}$ such that
$$
K_j\subset \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}\left(x_{j,l},\frac{\varepsilon}{2}\right)
$$
Since the series $\sum\limits_{j=1}^\infty\eta_j$ converges there exist $m\in\mathbb{N}$ such that $\sum\limits_{j=m}^\infty\eta_j<\frac{\varepsilon}{2}$. Then for all $j>m$
$$
\begin{align}
K_j&\subset K_m+\mathrm{Ball}(0,\eta_m)+\ldots+\mathrm{Ball}(0,\eta_{j-1})\\
&\subset K_m+\mathrm{Ball}\left(0,\sum\limits_{i=m}^j\eta_j\right)\\
&\subset K_m+\mathrm{Ball}\left(0,\sum\limits_{i=m}^\infty\eta_j\right)\\
&\subset K_m+\mathrm{Ball}\left(0,\frac{\varepsilon}{2}\right)\\
&\subset \bigcup\limits_{l=1}^{N_m}\mathrm{Ball}\left(x_{m,l},\frac{\varepsilon}
{2}\right)+\mathrm{Ball}\left(0,\frac{\varepsilon}{2}\right)\\
&\subset \bigcup\limits_{l=1}^{N_m}\left(\mathrm{Ball}\left(x_{m,l},\frac{\varepsilon}{2}\right)+\mathrm{Ball}\left(0,\frac{\varepsilon}{2}\right)\right)\\
&\subset \bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)
\end{align}
$$
Since the last inclusion holds for all $j>m$ we conclude
$$
\bigcup\limits_{j=m+1}^\infty K_j\subset \bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)
$$
Hence
$$
\begin{align}
\bigcup\limits_{j=1}^\infty K_j&=\left(\bigcup\limits_{j=1}^m K_j\right)\cup\left(\bigcup\limits_{j=m+1}^\infty K_j\right)\\
&\subset\left(\bigcup\limits_{j=1}^m \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}\left(x_{j,l},\frac{\varepsilon}{2}\right)\right)\cup\left(\bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)\right)\\
&\subset\left(\bigcup\limits_{j=1}^m \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}(x_{j,l},\varepsilon)\right)\cup\left(\bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)\right)\\
&=\bigcup\limits_{j=1}^m \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}(x_{j,l},\varepsilon)
\end{align}
$$
Thus, for each $\varepsilon>0$ we found finite $\varepsilon$-net $\{x_{j,l}:l=1,\ldots,N_j,\;j=1,\ldots,m\}$ for the set $\bigcup\limits_{j=1}^\infty K_j$. Hence it is relatively compact.