Consider e. g. $P(X)$, the space of probability measures over some compact metrisable space, $X$. This may be viewed as a WOT*-compact subspace of some dual Banach space $Y^{\ast}=(C(X),\|\cdot\|_{\infty})^{\ast}$, namely a certain closed subset of the unit ball: $\{\varphi\in Y^{\ast}:\|\varphi\|\leq1\wedge \varphi\geq 0\wedge\varphi(1)=1\}$. This is a compact polish space.
My question essentially has to do with such spaces. I now formulate the question a little abstractly, but keep in mind the above context.
Question: Let $Y$ be a Banach space. Consider a separable metrisable compact subset $B\subseteq Y^{\ast}$. Does $B$ possess a weak*-Schauder-Basis? That is, does there exist $(x_{n}^{\ast})_{n\in\omega}\subseteq B$, so that
$$\forall{x^{\ast}\in B:~}\exists{!(c_{n})_{n\in\omega}\subseteq\mathbb{C}:~} x^{\ast}=\sum_{n\in\omega}c_{n}x_{n}^{\ast}\quad(:=\text{WOT*-}\lim_{n}\sum_{k<n}c_{k}x_{k}^{\ast})$$
I know that not every separable Banach space has Schauder-basis. But here the situation is that of with separable subsets of (dual) Banach spaces viewed under WOT(*), and the search is for a weak${}^{\ast}$-basis.
The answer to the general question is no.
It is a result of Banach that a basis for the weak topology is also a basis for the norm topology. Now take a reflexive space $X$ without the approximation property (for instance, any subspace of $\ell_p$ where $p\neq 1,2,\infty$ which lacks the approximation property). The ball $B$ of $X$ is weakly* (=weakly) compact. It cannot have the property you want as otherwise $X$ would have a basis.
As for $P(X)$, have you tried Dirac deltas corresponding to points from a countable dense subset of $X$?