The Krein-Smulian Theorem states that for a convex set S, having weak*-closed intersections with closed balls implies being weak*-closed.
I would be really happy to have an example of a NOT weak-closed convex subset $S$ of a Banach space, such that the intersection with the closed balls are always weak-closed.
In this post, there are some counterexamples for the "convex" hypothesis. I would like to have counterexamples for the "weak*" hypothesis.
This is true for the weak topology too and it is easy to prove.
Proof. By the geometric form of the Hahn-Banach theorem, for a convex subset of $X$ to be closed is the same as to be weakly closed so it is enough to show that $S$ is closed.
Suppose that $S\subset X$ is convex and $S\cap B$ is (weakly) closed for every closed ball $B$ in $X$. Let $(x_n)_{n=1}^\infty$ be a sequence in $S$ which converges to some $x\in X$. Then $(x_n)_{n=1}^\infty$ is bounded, so it is contained in some ball $B$. However $S\cap B$ is closed, so $x\in S\cap B$. Consequently, $S$ is (weakly) closed. $\square$