[proofplan] We construct a metric on $B_{X^*}$ using a countable dense subset $\{x_n\}$ of $X$ and show it induces the weak* topology. The key ingredients are: the series $d(f,g) = \sum 2^{-n} |f(x_n) - g(x_n)|/(1 + |f(x_n) - g(x_n)|)$ converges and separates points (by density), $d$-convergence implies weak* convergence (by an $\varepsilon/3$ argument using density and uniform boundedness), and weak* convergence implies $d$-convergence (by controlling the tail of the series). [/proofplan] [step:Define the metric and verify it separates points] Let $\{x_n\}$ be a countable dense subset of $X$. Define \begin{align*} d(f, g) = \sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{|f(x_n) - g(x_n)|}{1 + |f(x_n) - g(x_n)|} \end{align*} for $f, g \in B_{X^*}$. Each summand is bounded by $2^{-n}$, so the series converges. [claim:Separation Of Points] $d(f, g) = 0$ implies $f = g$ on $B_{X^*}$. [/claim] [proof] If $f(x_n) = g(x_n)$ for all $n$, then $f - g$ vanishes on a dense subset. Since $f - g$ is a bounded linear functional (hence continuous), $f = g$ on all of $X$. [/proof] [/step] [step:Show $d$-convergence implies weak* convergence] [claim:Metric Convergence Implies Weak Star] If $d(f_k, f) \to 0$ with $f_k, f \in B_{X^*}$, then $f_k \overset{*}{\rightharpoonup} f$. [/claim] [proof] $d(f_k, f) \to 0$ implies $f_k(x_n) \to f(x_n)$ for each $n$. For arbitrary $x \in X$ and $\varepsilon > 0$, choose $x_n$ with $\|x - x_n\| < \varepsilon/4$. Then \begin{align*} |f_k(x) - f(x)| \leq \|f_k\|\|x - x_n\| + |f_k(x_n) - f(x_n)| + \|f\|\|x_n - x\| \leq \frac{\varepsilon}{2} + |f_k(x_n) - f(x_n)|, \end{align*} which is $< \varepsilon$ for $k$ large. [/proof] [/step] [step:Show weak* convergence implies $d$-convergence] [claim:Weak Star Implies Metric Convergence] If $f_k \overset{*}{\rightharpoonup} f$ in $B_{X^*}$, then $d(f_k, f) \to 0$. [/claim] [proof] Let $\varepsilon > 0$. Choose $N$ with $\sum_{n > N} 2^{-n} < \varepsilon/2$. Since $f_k(x_n) \to f(x_n)$ for $n = 1, \ldots, N$, the first $N$ terms of $d(f_k, f)$ are $< \varepsilon/2$ for $k$ large. [/proof] [/step] [step:Conclude metrizability] The previous two steps show that $d$-convergence and weak* convergence coincide on $B_{X^*}$. Since $d$ is a metric, $B_{X^*}$ is metrizable in the weak* topology. [/step]