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.
Hint: First assume that $\mathbf{E}$ is separable.
Theorem 1. Let $X$ be reflexive, $M$ a closed linear subspace of $X$. Then $M$ is reflexive.
Corollary 1. $X$ reflexive and separable $\iff$ $X'$ reflexive and separable.
My Proof:
Set $M = \overline{\mbox{span}_{\mathbb K}}\{x_n\}$. By Theorem 1, $M$ is reflexive and obviously is separable. Hence, by Corollary 1, $M'$ is reflexive and separable. Then $J(B_M) = B_{M''}$ is compact and metrizable with respect to the $\sigma(M'', M')$ topology, so there exists $\{x_{n_k}\}$ subsequence and $F = J(x)$ s.t. $J(x_{n_k}) \rightharpoonup J(x)$ when $k\to \infty$ with respect to the $\sigma(M'', M')$, i.e., by definition, $\langle J(x_{n_k}), f \langle_{M'',M'} \to \langle J(x_n), f \rangle_{M'',M'}$. Since $M$ is closed, $x \in M$ and $\langle f, x_{n_k} \rangle_{M',M} \to \langle f, x_n \rangle_{M',M}$ when $k\to\infty$. Thanks to the Hahn-Banach theorem, there exists $L \in X'$ s.t. $f := L_{|M}$ and hence we have $\langle L, x_{n_k} \rangle_{X',X} \to \langle L, x_{n_k} \rangle_{X',X}.$ $\square$
Is that right ? Or are there better solutions which are more detailed possible?