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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 2 / a / 3 / Solution

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A product of at most continuum many separable spaces is separable. More generally, if each factor has a dense subset of cardinality at most an infinite cardinal $\kappa$ and there are at most $2^\kappa$ factors, then the product has a dense subset of cardinality at most $\kappa$.


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 Incoming links: Hewitt–Marczewski–Pondiczery theorem