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.
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".