The standard answer I have seen on stackexchange is something like this: Hilbert space is the dual of itself, so applying Banach-Alaoglu, we see that the bounded sequence is contained in a closed ball which is weakly compact. Thus the sequence contains a weakly convergent subsequence.
However, weak compactness does NOT necessarily imply sequential weak compactness. This turns out to be true, by the Eberlein-Smulian Theorem, which states thtat weak compactness and sequential weak compactness are equivalent for Banach spaces. My question is: Is there any proof of the fact not using Eberlein-Smulian? Otherwise the urban legend that you only need Banach-Alaoglu to show this fact needs to be corrected!