 Solution

This statement is false. Let $I$ have the cardinality of the continuum and take $X={\ell }^1(I)$, which is nonseparable because its unit coordinate vectors form an uncountable discrete set. Its dual ball with the weak-star topology is the product cubeThe Hewitt–Marczewski–Pondiczery theorem says that a product of at most continuum many separable spaces is separable, so this cube is separable. Since $X^{\ast }={\bigcup }_{n\ge 1}n{B}_{X^{\ast }}$, the whole dual is weak-star separable despite $X$ being nonseparable.