Here is my effort to show this fact and I will use ball $X^*$ to denote the closed unit ball of $X^*$.
To show ball $X^*$ is weak star metrizable, we only have to show there is a metric $d$ on ball $X^*$ such that the topology induced by $d$ is the weak-star topology on ball $X^*$. Since $X$ is separable, ball $X$ is also separable. Thus there exists a countable dense subset $\{x_n\}$ in ball $X$. Now define the metric $d$ on ball $X^*$ by
$$d(x^*,y^*)=\sum_{n=1}^{\infty}\frac{|\langle x_n,x^*-y^*\rangle|}{2^n},$$
where $x^*,y^*\in\text{ball $X^*$}$. Let $T$ be the topology induced by $d$ and $wk^*$ be the weak-star topology on $\text{ball $X^*$}$. Then we need to show $T=wk^*$. And I try to use net convergence to show topology equivalence.
Let $x^*\in\text{ball $X^*$}$ and let $x_i^*$ be a net in $\text{ball $X^*$}$ such that $x_i^*\overset{wk^*}{\longrightarrow} x^*$. Then $\langle x_n,x_i^*\rangle\rightarrow\langle x_n,x^*\rangle$ for all $n$. Now Let $x^*\in\text{ball $X^*$}$ and let $x_i^*$ be a net in $\text{ball $X^*$}$ such that $x_i^*\rightarrow x^*$ in $(X,T)$. Then for each $\epsilon>0$, there exists $i_\epsilon\in\mathbb{N}$ such that $d(x_i^*,x^*)<\epsilon$ for all $i\geqslant i_\epsilon$; that is, for each $\epsilon>0$,
$$d(x_i^*,x^*)=\sum_{n=1}^{\infty}\frac{|\langle x_n,x_i^*-x^*\rangle|}{2^n}<\epsilon.$$
Here I want to show $x_i^*\overset{wk^*}{\longrightarrow} x^*$ if and only if $x_i^*\rightarrow x^*$ in $(X,T)$. But I forget some knowledge in Calculus. Can somebody help me to show this please? Thank you so much!!
Best Answer
Let $I$ denote the identity map from $(X^*,\text{weak*})$ onto $(X^*,d)$. Since $I$ is a bijection from a compact space (Banach-Alaoglu) into a Hausdorff space, we only need to show that $I$ is continuous.
To this end, fix a net $(x^*_\alpha)$ in $B_{X^*}$ such that $x^*_\alpha\to x^*$ in the weak* topology on $B_{X^*}$ for some $x^*\in B_{X^*}$. We want to show that $x^*_\alpha=I(x^*_\alpha)\to I(x^*)=x^*$ in the metric $d$; that is, we want: $$ \sum_{n=1}^\infty \frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} \longrightarrow 0. $$ To this end, fix $\varepsilon>0$. Since each $x_n$ is in the unit ball of $X$ and $x^*$ and each $x^*_\alpha$ are in the unit ball of $X^*$, we deduce $$ \frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} \le \frac{|x_\alpha^*(x_n)|+|x^*(x_n)|}{2^n} \le \frac{2}{2^n} = \frac{1}{2^{n-1}}. $$ Since $\sum_{n=1}^\infty2^{1-n}$ converges, there exists $N\in\mathbb{N}$ such that $\sum_{n=N+1}^\infty 2^{1-n}<\varepsilon/2$. Now, using the fact that $x_\alpha^*\to x^*$ in the weak* topology on $B_{X^*}$, for each $n\in\{1,\ldots,N\}$ there exists $\alpha_n$ in the directed set such that $$ |\langle x_n,x_\alpha^*-x^*\rangle| < \frac{2^n\varepsilon}{2N} $$ whenever $\alpha\ge\alpha_0$. Taking $\alpha_0$ in the direct set such that $\alpha_0\ge\alpha_1,\ldots,\alpha_N$, we obtain \begin{align*} \sum_{n=1}^\infty\frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} &= \sum_{n=1}^N\frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} +\sum_{n=N+1}^\infty\frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} \\ &\le \sum_{n=1}^N \frac{\varepsilon}{2N} + \varepsilon/2 \\ &= \varepsilon \end{align*} whenever $\alpha\ge\alpha_0$. This completes the proof.
A few comments on your post:
You write "Since $X$ is separable, ball $X$ is separable". This is a true statement, but only because $X$ is a metric space. In general topological spaces, subsets of separable spaces aren't necessarily separable.
It's important to show that $d$ is a metric. It is here that the density of $\{x_n : n\in\mathbb{N}\}$ in the unit ball of $X$ is actually needed.
You wrote "Then for each $\varepsilon>0$, there exists $i_\varepsilon\in\mathbb{N}$ such that $d(x_i^*,x^*)<\varepsilon$ for all $i\ge i_\varepsilon$." The problem is that $i_\varepsilon$ cannot be assumed to be a natural number, since your net may be indexed by some directed set other than $\mathbb{N}$.