Let $K \subset \ell^2(\mathbb{N})$ be a set defined as follows: $$ K := \left\{x = (x_1, x_2, \dots) \in \ell^2(\mathbb{N}) \,|\, |x_n| \le \frac{1}{n}\right\}.$$ Since $\ell^2(\mathbb{N})$ is a reflexive space, we know that any sequence of $K$, being bounded, admits a converging subsequence by the Eberlein-Smulian theorem. Moreover, since $K$ is closed, the weakly converging subsequence has a limit in $K$, and thus $K$ is weakly sequentially compact.
Since $(\ell^2(\mathbb{N}))^*$ is separable, we know that $\overline{B(0,1)}$ is metrizable for the weak topology and hence $K$ is not only weakly sequentially compact but compact for the weak topology.
Is it compact for the strong topology as well?
Yes it is. It is a standard (I mean "very common") argument in functional analysis, you can pick a sequence of points $(x_n)_n \in K^{\mathbb{N}}$ and proceed a Cantor's diagonal argument to extract a convergence subsequence. Then, it is just a matter of applying Lebesgue Dominated Convergence theorem to conclude convergece in $||.||_{\ell_2}$ as well.