Does anyone have a reference for the following result? I need to know the proof but I can't find it anywhere.
Lemma (Grothendieck, for weakly compact sets). Let $X$ be a Banach space, and $K \subset X$. If $K$ is weakly closed and bounded, and for all $\varepsilon >0$ there exists a weakly compact set $K_\varepsilon$ such that $K \subset K_\varepsilon + \varepsilon B_X$, then $K$ is weakly compact.
This lemma is in Grothendieck's book Espaces Vectoriels Topologiques, page 401.
The proof uses the bidual $X''$ whose unit ball is $\sigma(X'',X')$ compact by Alaoglu's theorem. Since $K$ is bounded its closure $\tilde K=\overline{K}^{\sigma(X'',X')}$ is $\sigma(X'',X')$-compact and it is enough to show $\tilde K\subseteq X$ (since the relative topology of $\sigma(X'',X')$ induced on $X$ is $\sigma(X,X')$ this then implies that $K$ is weakly relatively compact and hence weakly compact since $K$ was assumed to be weakly closed). For all $\varepsilon>0$ take a weakly compact $K_\varepsilon$ with $K\subseteq K_\varepsilon +\varepsilon B_X$. As the sum of a compact and a closed set is closed this implies $$\tilde K \subseteq K_\varepsilon +\varepsilon \overline{B_X}^{\sigma(X'',X')} \subseteq X+ \varepsilon B_{X''}.$$ Taking the intersection over all $\varepsilon>0$ yields that $\tilde K$ is contained in the norm-closure of $X$ in the bidual which is $X$ because $X$ is complete.