If a subspace of $X^*$ separates points, is it weak-* dense?
Solution 1:
Yes, it is weak-* dense. The weak-* continuous linear functionals on $X^*$ are, by definition, evaluation at the members of $X$. If $E$ was not weak-* dense in $X^*$, then by the separation theorem in topological vector spaces there would be such a functional that was $0$ on $E$ and not identically $0$. Since $E$ separates points, that is not the case.