Compactness in Hilbert space $\ell^2$

Question

If we have a compact set $A$ contained in $\ell^2$ (Hilbert Space), how can we show that $A$ is closed and bounded?

Answer 1

This holds in any metric space, closedness because of Hausdorffness and boundedness by considering covers like $\{B(x,n): m \in \mathbb{N}\}$ for a fixed $x$.

Closed and bounded does not imply compactness, the unit ball is a counterexample (what limit can a subsequence of $(e_n)$ have?).