Closed bounded sets need not be compact

Counterexample. In $\ell^2$ (space of square-summable sequences with $\|\cdot\|_2$), the closed unit ball $\overline B(0,1)$ is closed and bounded. But the sequence $(e_n)$ of standard basis vectors is at pairwise distance $\sqrt2$, so has no convergent subsequence.

Conclusion. In infinite dimension, closed bounded $\not\Rightarrow$ compact.

Note. Riesz's theorem: the closed unit ball of a normed space is compact iff the space is finite-dimensional.