Does separability follow from weak-* sequential separability of dual space?

Let $E$ be a Banach space. Suppose that $E'$ is weakly-* sequentially separable, that is, that there exists a countable $D \subset E'$ s.t. every $x' \in E'$ is a limit point of a sequence in $D$. Does it follow that $E$ is separable?

This question arises from a conversation with a friend of mine. He thinks this is true and plans to use it to prove separability of $C(K)$ for a metrizable and compact Hausdorff $K$. I'm not so sure this can work, though. Of course, if $E'$ is norm separable then $E$ is separable, but we are talking about a much weaker topology here.

Let $JT$ denote the James Tree space, $Q: JT \longrightarrow JT\,^{\prime\prime}$ the canonical embedding, $D_\ast$ a countable norm-dense subset of $JT$, let $D= Q(D_\ast)$ and let $E = JT\,^\prime$. Note that $E$ is nonseparable in the norm topology and that the $w^\ast$-sequential closure of $D$ in $E^{\prime}$ is $E^{\prime}$ since $Q(JT)$ is $w^\ast$-sequentially dense in $JT\,^{\prime\prime}$ (for the last claim, see in particular Corollary 2 of Lindenstrauss and Stegall Examples of separable spaces which do not contain $\ell_1$ and whose duals are non-separable, Studia Math. 54 (1975), p.81--105).

I should mention that until the appearance of the Lindenstrauss-Stegall result cited above, it seems to have been open since the time of Banach whether there could exist a separable Banach space that has nonseparable dual and is $w^\ast$-sequentially dense in its bidual.