Can we classify weakly compact subsets of $l^{\infty}$? My question is as stated. The motivation behind this is that I want to prove that such subsets are norm-separable.

Edit: my previous answer was plain wrong as the set $B_K$ need not be a ball of any space (for example, take the unit vector basis of $c_0$; $c_0$ does not contain any infinite-dimensional reflexive subspaces). Here is an easier argument.
Suppose that $K\subset \ell_\infty$ is weakly compact. As $\ell_\infty$ is isometric to the dual space of $\ell_1$, it carries the weak* topology, which is coarser than the weak topology of $\ell_\infty$. In particular, $K$ is also compact in the weak* topology and the two relative topologies agree on $K$. However, as $\ell_1$ is separable, weak* compact subsets of $\ell_\infty$ are second-countable. One may use Mazur's lemma (as explained here) to conclude that a set is weakly separable if and only if it is norm-separable.
Suppose that $K\subset \ell_\infty$ is weakly compact. Then, by Mazur's lemma, so is the closed convex hull of $K\cup \{0\}$, call it $B_K$. Consider $\ell_\infty$ as the dual space to $\ell_1$. Since the weak* topology on $\ell_\infty$ is coarser than the weak topology, by compactness, they must be equal on $B_K$. Since $B_K$ is weakly compact, $X_K$ is reflexive. However, the weak*-topology and the weak topologies of $\ell_\infty$ coincide on $X_K$, so $X_K$ is weak*-closed. Then, by reflexivity, $X^*_K$ is isomorphic to $\ell_1 / (X_K)_\perp$, where $(X_K)_\perp$ is the pre-annihilator of $X_K$ so $X_K$ (and hence $K$ itself) is norm-separable. 
This can be generalised as follows. If $\Gamma$ is infinite, then each weakly compact subset of $\ell_\infty(\Gamma)$ has density at most $|\Gamma|$. (For example, for regular cardinals, it follows from Theorem 2.6 here.)
