= 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 cube
$$
B_{X^*}=[-1,1]^I.
$$
The <Hewitt–Marczewski–Pondiczery theorem> says that a product of at most continuum many separable spaces is separable, so this cube is separable. Since $X^*=\bigcup_{n\geq1}nB_{X^*}$, the whole dual is weak-star separable despite $X$ being nonseparable.
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