Let $X$ be a Banach space and $W$ its weakly compact subset. Let $(x_n)$ be e sequance of elements from convex hull of $W$. We can deduce that each such sequance is in a separable subspace of X. Indeed, let $S=span_{\mathbb{R}}\{(x_n)\}$. Then $span_{\mathbb{Q}}\{(x_n)\}$ is countable and dense in $S$, so $S$ is separable. Can we deduce somehow from Hahn-Banach theorem, that we may assume $W$ to be separable (without a loss of generality)?
Motivation.
Suppose (somehow) that $W$ is norm separable in Banach space $X$.Define $g \colon W\to W$ by
$$g(x)=x.$$
If $W$ is separable then so is $g(W)$. This means that $g$ is separably valued. Take any $x^*\in X^*$. Then, if $x_n\to x$, we have
$$\langle x^*,g(x_n)\rangle\to\langle x^*, g(x)\rangle.$$ Now, from Pettis measurablity theorem it follows that $g$ is $\mu$-measurable for every regular (countable additive) measure $\mu$ on the weak Borel sets. From Kakutani representation theorem, we know that for every such measure $\mu$ there is a unique linear and continuous functional $\psi$ such that
$$\psi(f)=\int_{W}f\,d \mu$$
for every $f\in C_{0}(W)$. Hence, we think about $\mu$ that it is an element of $C_{0}^*(W)$. Define an operator $T\colon C^*_{0}(W)\to X$ by
$$T(\mu)=\int_{W}g\,d\mu.$$
This opearor is well defined ($W$ is weakly compact, so $W$ is norm bounded). Let us endow $C^*_0(W)$ with weak$*$ topology and let $\mu_n\to \mu$. Take any $x^*\in X^*$. It follows from characterization of $C^*_0(W)$ endowed with weak$*$ topology that
$$\lim_{n\to \infty}x^*T(\mu_n)=x^*(\lim_{n\to \infty} \int_{W}g\,d\mu_n)=x^*T(\mu).$$
Now, we can deduce that operator $T$ is continuous from $C^*_{0}(W)$ endowed with weak$^*$ topology to $X$ with weak topology. Let $S^*$ be unit ball of $C^*_0(W)$. Then from Banach-Alaoglu theorem we deduce that $S^*$ is compact in $C^*_0(W)$ endowed with weak$^*$ topology. From the continuity of $T$ we obtain that $T(S^*)$ is weakly compact in $X$. The conclusion is that operator $T$ is "weakly" compact (with suitable topologies). The compactness of operator $T$ allows to deduce that the closed convex hull of $W$ is weakly compact.
Best Answer
Your proof has some really interesting ideas. However, you cannot assume that $W$ is separable or even weakly separable by Hahn-Banach. The closed unit ball of any non-separable Hilbert space is weakly-compact by reflexivity and Banach-Alaoglu, but it is not weakly separable (assume to the contrary that it is, and use the fact the weak closure of convex set is the same as the norm closure to derive a contradiction), and definitely not norm separable.
I think you are on the right track though. For convenience let $H$ be the closed convex hull of $W$. We know via the Eberlein-Smulian Theorem that weak compactness is equivalent to sequential weak compactness. Thus let $(x_n)\subset H$. As $H$ is the closed convex hull we have that each $x_n=\lim_{m\to \infty}z_m$, where each $z_m$ is a convex combination of points in $W$. Assuming the axiom of choice we thus have a countable set $W_1\subset W$ such that the closed convex hull of $W_1$, say $H_1$, contains $(x_n)$. Now $H_1$ is separable in a metrizable space, so the weak closure of $W_1$ must also be separable. The weak closure of $W_1$ is also weakly compact because $W$ is weakly compact. Thus if we can show that $H_1$ is weakly compact we will have the existence of a convergent subsequence of $x_n$, which implies that $H$ is compact.
It is in this manner we can assume that $W$ is separable I believe, by actually using your argument on the weak closure of $W_1$ to prove that $H_1$ is weakly compact.