Let $X$ be a separable Banach space and let $(x_n)_{n \in \mathbb{N}}$ be a fixed sequence such that $x_n \neq 0$ for all $n \in \mathbb{N}$ and $\lbrace x_n: n\in \mathbb{N}\rbrace$ is dense in $X$. Moreover, let $X^\prime$ be the dual space of $X$ and $M=\overline{B_1(0)} \subset X^\prime$ the closed unit ball in $X^\prime$ w.r.t. $||\cdot ||_{X^\prime}$. For $x^\prime, y^\prime \in M$ define
$$d(x^\prime,y^\prime)= \sum\limits_{n=1}^{\infty} \frac{1}{2^n||x_n||_X}|x^\prime(x_n)-y^\prime(x_n)|,$$ which is a metric on $M$. Given a sequence $(x_m^\prime)_{m \in \mathbb{N}} \subset M$ and $x^\prime \in M$, I have to show that $$x_m^\prime \longrightarrow x^\prime \text{ weakly* as } m \longrightarrow \infty \text{ iff } d(x_m^\prime,x^\prime) \longrightarrow 0 \text{ as }m \longrightarrow \infty.$$
I wanted to show the statement in two parts:
1) $d(x_m^\prime,x^\prime) \longrightarrow 0 \Longleftrightarrow x_m^\prime(x_n)-x^\prime(x_n) \longrightarrow 0 \text{ as } m \longrightarrow \infty \forall n \in \mathbb{N}$
2) $x_m^\prime \longrightarrow x^\prime \text{ weakly* as } m \longrightarrow \infty \Longleftrightarrow x_m^\prime(x_n)-x^\prime(x_n) \longrightarrow 0 \text{ as } m \longrightarrow \infty \forall n \in \mathbb{N}$
Part 2) and one implication in part 1) were easy to prove, but I still struggle with showing $ x_m^\prime(x_n)-x^\prime(x_n) \longrightarrow 0 \text{ as } m \rightarrow \infty \forall n \in \mathbb{N} \Longrightarrow d(x_m^\prime,x^\prime) \longrightarrow 0 $ as $m \rightarrow \infty$.
Can we conclude that $(M,d)$ is a compact metric space by Banach Alaoglu?
Let $\epsilon > 0$. We want to show that there exists a number $m_0$ (I wish $M$ were not already used) such that for all $m \ge m_0$ we have $d(x_m',x') < \epsilon$.
First note that $$\frac{1}{2^n||x_n||_X}|x_m^\prime(x_n)-x^\prime(x_n)| \le \frac{2}{2^n}$$ and so if we choose $N$ so large that $\sum_{n={N+1}}^\infty \frac{2}{2^n} < \epsilon/2$, then it suffices to show that $$\sum_{n=1}^N \frac{1}{2^n||x_n||_X}|x_m^\prime(x_n)-x^\prime(x_n)| < \frac{\epsilon}{2}.$$ In particular, it would suffice to have $$\frac{1}{2^n||x_n||_X}|x_m^\prime(x_n)-x^\prime(x_n)| \le \frac{\epsilon}{2} \tag{*}$$ for every $n=1,\dots,N$. Now since $x_m^\prime(x_n) \to x^\prime(x_n)$ for every $n$, you can choose $m_0$ large enough so that (*) holds for every $n=1,\dots, N$ and every $m \ge m_0$.
This is a pretty standard technique for showing convergence of infinite sums with respect to a parameter: you "cut off" the sum at some point such that the tails are uniformly small, and thus you reduce the problem to the convergence of a finite sum, which is much easier.
As to your last comment: yes, you know by Banach-Alaoglu that the weak-* topology on $M$ is compact. So when you show that the metric $d$ induces the weak-* topology on $M$, then indeed you know that $(M,d)$ is a compact metric space.