Unit ball in dual space is weak*separable

Question

Let $X$ be a separable Banach space, $X'$ the dual space of $X$ endowed with the weak$*$-topology. How to prove that the unit ball $B$ in $X'$ is weak*-separable ?

I only know Banach-Alaoglu, which states that $B$ is compact in the weak*-topology, but not sure if it helps.

Answer 1

Use this question to show that

Theorem. If X is separable, then $B$ endowed with the weak*-topology is metrizable.

Since you also know that $B$ is compact, all you need now is contained in that question:

Theorem. If $X$ is a compact metric space, then $X$ is separable.

Answer 2

Let $x_n$ be a countable dense subset of the unit ball $B_X = \{x \in X: \|x\| \le 1\}$ of $X$. Then if $f \in B_{X'}$, the unit ball of $X'$ in the weak$^\ast$ topology, we define $F(f) = (f(x_n))_n \in D^\mathbb{N}$ where $D = \{x \in \mathbb{F}: \|x\| \le 1\}$, where $\mathbb{F}$ is the scalar field (reals or complex numbers), so $D$ is compact. Then in this answer I show that $F$ is 1-1 and so as $B_{X'}$ is compact, $F$ is an embedding and so $B_{X'}$ is metrisable and being compact separable as well.