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> Let $\mathbf{E}$ be a reflexive space and $\mathbf{A ⊂ E}$ be bounded and weakly closed. Show that $\mathbf{A}$ is sequentially compact, i.e., every sequence in $\mathbf{A}$ has a weakly convergent subsequence with limit in $\mathbf{A}$. This statement is a special case of the Eberlein-Smulian Thoerem, which you are not allowed to use in this task.
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Hint: First assume that $\mathbf{E}$ is separable.

I just can solve it with this Eberlein Smulian Theorem and I think my other second approach is wrong... Has anybody a solution with an explication, why I should assume, that $\mathbf{E}$ is separable.

I also saw a proof of this here [Every bounded sequence has a weakly convergent subsequence in a Hilbert space](https://math.stackexchange.com/questions/1177782/every-bounded-sequence-has-a-weakly-convergent-subsequence-in-a-hilbert-space) , but there they start with a Hilbert space? Hilbert space is reflexive but not the other way. Or is there no connection?