It is known that for a metrizable space $(X,\vert \vert \cdot \vert \vert)$ where $\vert \vert \cdot \vert \vert$ is a norm on $X$, that:
sequential compactness $\iff$ compactness
And further for the weak topology on $X$, if $X$ is Banach space then:
weak sequential compactness $\iff$ weak compactness
I am struggling to find an intuition as to why we need the additional condition of $X$ Banach.
My thinking: Since by definition $\tau_{\operatorname{weak}} \subseteq \tau_{\operatorname{strong}}$ in terms of topologies, surely the condition of weak sequential compactness being equivalent to weak compactness is a lot "easier" than showing the same for a finer topology $\tau_{\operatorname{strong}}$?
I mean surely if $X$ compact then $X$ has to be weakly compact, given the coarser topology of $\tau_{\operatorname{weak}}$
Consider any set $X$ with two topologies $\tau, \tau'$. If $\tau$ is coarser than $\tau'$, then obviously the following is true:
If $C$ is a compact subset of $(X, \tau')$, then it also a compact subset of $(X, \tau)$.
Consider any open cover $\mathcal U$ of $C$ in $(X, \tau)$. Then $\mathcal U$ is also an open cover in $(X, \tau')$ and thus has a finite subcover.
Be aware that in general there will be compact subsets of $(X, \tau)$ which are not compact in $(X, \tau')$.
Similarly it is obviuos that each convergent sequence in $(X, \tau')$ is also convergent in $(X, \tau)$ (with the same limit).
However, in your case the non-trivial theorem is the equivalence of weak sequential compactness and weak compactness. You cannot expect to get that for free.