This question could sound naive, but I honestly did not found anything around the Net (just some related topics on https://mathoverflow.net/q/177806/95288).
Consider the Banach space $\ell^\infty([0,1]):=L^\infty([0,1],2^{[0,1]},\#)$, where $\#$ is the counting measure, and its closed (both in norm and weak topologies) subspace $X:=\{x\in\ell^\infty([0,1])\colon\text{ spt}(x)\text{ at most countable}\}$. Since $\ell^\infty([0,1])$ can be identified with $\ell^1([0,1])^*$ by the Riesz representation, we could even consider the weak$^*$ topology on $\ell^\infty([0,1])$. Is $X$ weakly$^*$ closed?
I'm quite sure it is sequentially weakly$^*$ closed, since the weak$^*$ limit of a sequence of countably supported functions is still countably supported. Yet, it does not suffice to infer its topological weak$^*$ closedness: $\ell^1([0,1])$ is non-separable hence any chance of metricising the weak$^*$ topology on the unit ball of $\ell^\infty([0,1])$ and exploiting Krein-Smulian vanishes.
In general, how can one say anything about the topological weak$^*$ closedness of a set in the non-separable context?
Best Answer
In the weak$^*$ topology, $X$ is dense. To see this, let $z\in \ell^\infty[0,1]$. A weak$^*$-neighbourhood of $z$ is of the form $$ N=\{y\in\ell^\infty[0,1]:\ |\langle (y-z),w_j\rangle|<1,\ j=1,\ldots,m\}. $$ where $w_1,\ldots,w_m\in\ell^1[0,1]$. Let $M=\bigcup_{j=1}^m\operatorname{supp}w_j$, and let $x=z|_M$ (that is, $x(t)=z(t)$ if $t\in M$, and $0$ otherwise). Then $x\in X$ and $x\in N$. This shows that there exists a net $\{x_\alpha\}\subset X$ such that $x_\alpha\to z$ in the weak$^*$-topology.
The key, as shown above, is to think in terms of neighbourhoods. This is important because in general we have no obvious grounds to construct a net. The net $\{x_\alpha\}$ is indexed by the neighbourhoods. I'm not sure if anything else can be said in general.
(for anyone who knows a bit about operators)
The kind of unintuitive phenomena as above (where the use of nets is crucial to convergence) is similar to what happens with the weak operator topology in $B(H)$. Like the fact that the unitaries are dense in the whole unit ball, or the projections are dense in the set of positive operators in the unit ball, while a strong operator topology limit of unitaries or projections is respectively a unitary or a projection.