Let $l^{2}$ be the space of all square-summable sequences. Let $a>0.$ Consider the following subset of $l^{2}$: $$ S = \{\mathbf{x} = (x_{1}, x_{2}, \dots, ): \sum_{i}x_{i} = a, \,\,\, \text{and} \,\,\, x_{1} \geq x_{2} \geq \dots \}. $$
Is $S$ compact?
No, $S$ is not strongly compact and not even weakly compact, because it is a hyperplane defined by an unbounded (not-closed, not everywhere defined) linear functional and hence not even weakly closed.
In more detail: without restriction of generality, let $a=1$. For any ${\mathbf{x}}$ in $S$, if there were any negative $x_i$ it would make the infinite series $\sum_{i}x_{i}$ diverge, taking the boundary condition into account. So the $x_i$ can only be nonnegative. Then $\sum_{i}x_{i}^2 \leq \sum_{i}x_{i} = 1$, hence $S$ is a subset of the unit ball in the space of square-summable sequences. The unit ball is weakly metrizable, so it suffices to find a Cauchy sequence $(\mathbf{x_i})$ without a limit in $S$. Let the sequence $(\mathbf{x_i}) \subset S$ be
$$ \mathbf{x_1} = (1, 0, \dots ) $$ $$ \mathbf{x_2} = (1/2, 1/2, 0, \dots ) $$ $$ \mathbf{x_3} = (1/3, 1/3, 1/3, 0, \dots ) $$
and so forth. The sequence converges strongly and weakly to the zero vector which is not in $S$, so $S$ is not a closed subset of Hilbert space. Compact sets in Hausdorff spaces must be closed, so $S$ cannot be compact.