If $X$ is separable, then $B_{X^*}$ is metrizable for the weak$^*$ topology The following theorem and proof are extracted from 'Topics in Banach Space Theory', page $17$.

Let $X$ be a Banach space. If $X$ is separable, then $B_{X^*}$ of $X^*$ is (compact and) metrizable for the weak$^*$ topology. 

Proof: Let us take $(x_n)$ dense in the unit ball $B_X$ of $X$. We define the topology $\rho$ induced on $X^*$ by convergence on each $x_n$. Precisely, a base of neighbourhood for $\rho$ at a point $x_0^* \in X^*$ is given by the sets of the form 
$$V_{\varepsilon}(x_0^*: x_1,...,x_N) = \{ x^* \in X^*: |x^*(x_n) - x_0^*(x_n) | < \varepsilon , n =1,...,N\}$$
where $\varepsilon >0$ and $N \in \mathbb{N}$. This topology is metrizable, and a metric inducing $\rho$ may be defined by 
$$d(x^*,y^*) = \sum_{n=1}^{\infty}2^{-n} \min (1, |x^*(x_n) - y^*(x_n)|), x^*, y^* \in X^*.$$
The topology $\rho$ is Hausdorff and weaker than the weak$^*$ topology, so it coincides with the weak$^*$ topology on the weak$^*$ compact set $B_{X^*}$.
Questions:
$(1)$ How to show $d$ satisfies triangle inequality? Is the following inequality correct? 
$$min (1, |x^*(x_n) - y^*(x_n)|) \leq min(1, |x^*(x_n) - z^*(x_n)|) + min(1, |z^*(x_n) - x^*(x_n)| )?$$
$(2)$ Why the topology $\rho$ is Hausdorff and weaker than the weak$^*$ topology?
$(3)$ What is the significance of this theorem? 
 A: (1) Your inequality is right, we can prove it by dividing some cases. First:
$\min\{1,|x^*(x_n)-y^* (x_n)|\}=\min\{1,|x^*(x_n)-z^*(x_n)+z^*(x_n)-y^* (x_n)|\}\le\min\{1,|x^*(x_n)-z^*(x_n)|+|z^*(x_n)-y^* (x_n)|\}$.
Now if $\min\{1,|x^*(x_n)-z^*(x_n)|+|z^*(x_n)-y^* (x_n)|\}=1$, then:
$\min\{1,|x^*(x_n)-z^*(x_n)|+|z^*(x_n)-y^* (x_n)|\}\le \min\{1,|x^*(x_n)-z^*(x_n)|\}+\min\{1,|z^*(x_n)-y^* (x_n)|\}$
in fact if on the right there is at least a $1$ we are done, otherwise on the right there is $|x^*(x_n)-z^*(x_n)|+|z^*(x_n)-y^* (x_n)|$ that is by hypothesis greater or equal to $1$.
On the other hand if $\min\{1,|x^*(x_n)-z^*(x_n)|+|z^*(x_n)-y^* (x_n)|\}=|x^*(x_n)-z^*(x_n)|+|z^*(x_n)-y^* (x_n)|$, then $|x^*(x_n)-z^*(x_n)|$ and $|z^*(x_n)-y^* (x_n)|$ are both $\le1$, so the inequality is again verified (with $=$).
(2) Let us denote by $\tau_d$ the topology given by the distance $d$ on $B_{X^*}$ and let $\rho$ be the topology given by the neighbourhoods of the form $V_\epsilon\{x^*_0;x_1,\dots,x_N\}$. Let us call $B(x^*,r)$ a ball in the distance $d$ with center $x^*$ and radius $r$.
First: open sets of $\tau_d$ are open sets of $\rho$. In fact take a ball $B(x^*_0,r)$; for $x^*,y^*\in V_\epsilon\{x^*_0;x_1,\dots,x_N\}$ with $\epsilon<1$, since $|x^*(x_n)-y^*(x_n)|\le 2$, we have
$d(x^*,y^*)\le\epsilon\sum_{n=1}^{N}\frac{1}{2^n} +\sum_{n=N+1}^{+\infty}\frac{1}{2^{n-1}}$
that is $<r$ if $N$ is chosen sufficiently big and $\epsilon$ is chosen sufficiently small.
Second: the evaluation maps $J_k:f\mapsto f(x_k)$ are continuous with respect to the topology $\tau_d$. In fact let $U\subseteq \mathbb{R}$ open and take the preimage $J_k^{-1}(U)\ni f$; assume $(f(x_k)-\delta,f(x_k)+\delta)\subseteq U$. If $g\in B(f,r)$, then
$|f(x_k)-g(x_k)|=\min\{1,|f(x_k)-g(x_k)|\}< r 2^k$
that is $<\delta$ for $r$ sufficiently small. Hence $J_k^{-1}(U)$ is open, and $J_k$ is continuous in the topology $\tau_d$.
Third: if $x\in X$, the evaluation maps $J_x:f\mapsto f(x)$ are continuous with respect to the topology $\tau_d$. In fact let $U\subseteq \mathbb{R}$ open and take the preimage $J_x^{-1}(U)\ni f$; assume $(f(x)-\delta,f(x)+\delta)\subseteq U$. By density, there exists a sequence $x_k\to x$ in $X$, and thus $f(x_k)\to f(x)$; take $\bar{k}$ such that $f(x_{\bar{k}})\in(f(x)-\delta/4,f(x)+\delta/4)$ and $||x-x_{\bar{k}}||_{X}<\delta/4$. By the calculation of the second point there exists a ball $B(f,r)$ such that for all $g\in B(f,r)$ we have $g(x_{\bar{k}})\in(f(x_{\bar{k}})-\delta/4,f(x_{\bar{k}})+\delta/4)$; moreover $|g(x)-g(x_{\bar{k}})|\le ||g||_{X^*}||x-x_{\bar{k}}||_{X}<\delta/4$. Finally for all $g\in B(f,r)$:
$|g(x)-f(x)|\le |g(x)-g(x_{\bar{k}})| + |g(x_{\bar{k}})-f(x_{\bar{k}})|+|f(x_{\bar{k}} -f(x)|<3\delta/4$
Hence $J_x^{-1}(U)$ is open, and $J_x$ is continuous in the topology $\tau_d$.
Four: topology $\rho$ is contained in weak* topology $\tau$. In fact a generic neighbourhood in $\tau$ is
$W_\eta\{x_0^*;y_1,\dots,y_M\}=\{x^*: |x^*_0(y_i)-x^*(y_i)|<\eta, i=1,\dots,M \}$
If $N$ is sufficiently big, for all $i=1,\dots,M$ there is $j_i\in\{1,\dots,N\}$ such that $||x_{j_i}-y_i||_X<\eta/4$. So if $x^*\in V_{\eta/2}\{x^*_0;x_1,\dots,x_N\}$ then
$|(x^*-x^*_0)(y_i)|\le ||x^*-x^*_0||_{X^*}||y_i-x_{j_i}||_X+|(x^*-x^*_0)(x_{j_i})|<\eta$
Hence $V_{\eta/2}\{x^*_0;x_1,\dots,x_N\}\subseteq W_\eta\{x_0^*;y_1,\dots,y_M\}$ and thus $\rho\subseteq\tau$.
Five: By definition the weak* topology is the coarsest topology for which the evaluation maps are continuous; since $\tau_d\subseteq \rho\subseteq\tau$, then $\tau_d=\rho=\tau$ and we have the thesis.
(3) This theorem is very important, in fact with a metrizable topology, we in particular have a first countable topology. In this setting a set is closed if and only if it is sequentially closed; and a compact set is sequentially compact. In particular, since by Banach-Alaoglu $B_{X^*}$ is weak* compact, in this setting the ball is also sequentially compact; this is a property that often turns out to be very useful.
