Theorem:
Let $B$ be the closed unit ball of a Banach space $X$. Then $B$ is weakly compact if and only if it is weakly sequentially compact.
Forward direction:
We first assume $B$ is compact. Kakutani's Theorem tells us that $X$ is reflexive.
Every bounded sequence in $X$ has a weakly convergent subsequence. Since $B$ is weakly closed, $B$ is weakly sequentially compact.
Why can we assume $B$ is weakly closed? I'm guessing this follows from $B$ being strongly closed but I'm unclear on the relationship of the two.
I think weak closure implies strong closure: $B$ is weakly closed so its compliment is weakly open. Thus $B^C$ is certainly open in a stronger topology – such as the norm topology. Thus $B$ is closed in the norm topology. I don't see how the other direction can be true though because we are "coming from a finer topology".
Best Answer
$B$ weakly compact implies $B$ weakly closed, merely by $X$ being a Hausdorff space in the weak topology (which actually holds because we can separate points by functionals, Hahn-Banach etc.). This is just a special case of "compact sets are closed in Hausdorff spaces"! (general topology is actually useful, see?)
From this question (and general Banach space theory) we see that a strongly closed and convex subset is in fact weakly closed too. This applies to the closed unit ball $B$; if you already know the result you could use that as well. Of course, I like the general topology approach better, as it uses less.