Krein-Smulian counterexample for the weak (not *) topology.

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.

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. $\square$
• So, my conclusion is that the difficulty in proving Krein-Smulian rests in the fact that when you have a strongly closed convex set $S \subset X^*$ and a point $\phi \in X^*$, it is not as easy to find $x \in X$ separating $S$ and $\phi$, as it is easy to apply Hahn-Banach and get an $F \in X^{**}$. – André Caldas May 13 '17 at 10:26