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.

Another way to prove this is to use the following result:

Lemma: If $K \subset \ell_{\infty}$ is weakly compact, then it is weakly metrisable.
Proof: For $x,y \in K$ we set,
$$ d(x,y) = 2^{-n} |x_n-y_n|. $$
Since $K$ is norm bounded (by the uniform boundedness principle) the sum is finite and we can readily check this defines a metric on $K.$ Moreover it is a locally convex space defined by the coordinate maps $x \mapsto x_n.$
Since $(K,d)$ is generated by fewer functionals compared to the weak topology, it is coarser and so the identity map,
$$ (K,w) \rightarrow (K,d) $$
is continuous. Since $(K,w)$ is compact and $(K,d)$ is Hausdorff the map is a homeomorphism, so the topologies coincide.

Hence it follows that $K$ is weakly separable, by using total boundedness and taking $\frac1n$-nets.
Now let $\{x_n\}_{n=1}^{\infty}$ be a weakly dense subset of $K$ and set,
$$ C = \left\{ \sum_{i=1}^n t_i x_{k_i} \mid n \in \mathbb N, t_1, \dots, t_n \in [0,1] \cap \mathbb Q, \sum_{i=1}^n t_i = 1, k_1 < \dots < k_n \right\}. $$
That is, the rational convex hull of the $x_n$'s. One can easily show this is weakly dense in $\mathrm{conv}\ K,$ so by Mazur's lemma,
$$ \overline{\mathrm{conv}}^{\lVert\cdot\rVert} C = \overline{\mathrm{conv}}^w C = \overline{\mathrm{conv}}\ K. $$
Since $C$ is countable, $\overline{\mathrm{conv}}\ K$ and hence $K$ is norm separable.
