Corollary of Eberlein Smulian theorem

Question

I've read Conway's chapter 4-13.

An Eberlein-Smulian Theorem says that if $X$ is a Banach space and $A \subseteq X$, then TFAE.

1. Each sequence of elements of $A$ has a weakly convergent subsequence
2. Each sequence of elements of $A$ has a weak cluster point
3. The weak closure of $A$ is weakly compact

Next, the author leave me to prove the following corollary: If $X$ is a Banach space and $A \subseteq X$, then $A$ is weakly compact iff $A\cap M$ is weakly compact for every separable norm closed subspace $M$ of $X$.

The only if part is easy, since $M$ is convex so weak closed, and so $A\cap M$ is weakly compact.

For the if part, if we assume $A$ is weakly closed, then it sufficies to show every sequence in $A$ has a weakly convergent subsequence, by the Eberlein-Smulian Theorem. Now, if $x_n \in A$, then taking $M = \bar{span(x_n)}^{norm}$, we complete the proof.

However, I can't show $A$ is weakly closed. Can anyone help me to prove this ?

Thank you.

**Edit: By the above span - norm closure argument, we can see $A$ is weakly sequentially compact(That is, every sequence in $A$ has a convergent subsequence whose limit is in $A$). Thus it sufficies to show weakly sequentially compact implies weakly closed. (I already know that weakly sequentially 'closed' does not implies weakly closed).