[proofplan] We define a metric $d$ on $B_R$ using a countable dense subset of $X$ (which exists by separability), verify it is a metric, then show the $d$-topology coincides with the weak* topology on $B_R$ by proving both directions of convergence equivalence. The forward direction uses density and the uniform bound $R$ to approximate arbitrary evaluations; the reverse uses the fact that convergence at a dense set plus uniform boundedness implies convergence everywhere. [/proofplan] [step:Define the metric and verify non-degeneracy] Let $\{x_k\}_{k=1}^\infty \subseteq X$ be a countable dense subset (exists by separability). Define \begin{align*} d(f, g) &:= \sum_{k=1}^{\infty} 2^{-k} \frac{|f(x_k) - g(x_k)|}{1 + |f(x_k) - g(x_k)|} \end{align*} for $f, g \in B_R$. Symmetry and the triangle inequality follow from properties of $t \mapsto t/(1+t)$. Non-degeneracy: if $d(f, g) = 0$, then $f(x_k) = g(x_k)$ for all $k$. The functional $f - g \in X^*$ is continuous and vanishes on the dense set $\{x_k\}$, so $f = g$ on all of $X$. [/step] [step:Show $d$-convergence is equivalent to weak* convergence on $B_R$] ($\Rightarrow$) Suppose $d(f_\alpha, f) \to 0$. Then $f_\alpha(x_k) \to f(x_k)$ for each $k$. For arbitrary $x \in X$ and $\varepsilon > 0$, choose $x_k$ with $\|x - x_k\|_X < \varepsilon/(3R)$. Then \begin{align*} |f_\alpha(x) - f(x)| &\leq 2R\|x - x_k\|_X + |f_\alpha(x_k) - f(x_k)| < \frac{2\varepsilon}{3} + |f_\alpha(x_k) - f(x_k)|. \end{align*} Since $f_\alpha(x_k) \to f(x_k)$, eventually the second term is less than $\varepsilon/3$. ($\Leftarrow$) Suppose $f_\alpha(x) \to f(x)$ for every $x$. For any $\varepsilon > 0$, choose $K$ with $\sum_{k > K} 2^{-k} < \varepsilon/2$. For sufficiently large $\alpha$, the partial sum $\sum_{k=1}^{K}$ is less than $\varepsilon/2$. Hence $d(f_\alpha, f) < \varepsilon$. [/step]