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.