I'm reading the proof of Theorem 3.28. in Brezis's Functional Analysis, i.e.,
Theorem 3.28. Let $E$ be a separable Banach space. Then $B_{E^{\star}}$ is metrizable in the weak$^{\star}$ topology $\sigma\left(E^{\star}, E\right)$.
The author has a remark just right after the proof of Theorem 3.28, i.e.,
Remark 20. One should emphasize again (see Remark 3) that in infinite-dimensional spaces the weak topology $\sigma\left(E, E^{\star}\right)$ (resp. weak ${ }^{\star}$ topology $\sigma\left(E^{\star}, E\right)$ ) on all of $E$ (resp. $E^{\star}$ ) is not metrizable; see Exercise 3.8. In particular, the topology induced by the norm [ ] on all of $E^{\star}$ does not coincide with the weak ${ }^{\star}$ topology.
Below is the author's proof. In $\color{blue}{(\star)}$ and $\color{blue}{(\star\star)}$, we need $f-f_0$ to be bounded (here by $2$). The proof does not work if the subset we consider (here $B_{E^\star}$) is unbounded. Does the theorem hold if $B_{E^{\star}}$ is replaced by a non-empty bounded (in norm) subset of $E^*$.
Proof of Theorem 3.28. Let $\left(x_n\right)_{n \geq 1}$ be a dense countable subset of $B_E$. For every $f \in E^{\star}$ set $$ [f]=\sum_{n=1}^{\infty} \frac{1}{2^n}\left|\left\langle f, x_n\right\rangle\right| . $$
Clearly, $[ \cdot ]$ is a norm on $E^{\star}$ and $[f] \leq\|f\|$. Let $d(f, g)=[f-g]$ be the corresponding metric. We shall prove that the topology induced by $d$ on $B_{E^{\star}}$ is the same as the topology $\sigma\left(E^{\star}, E\right)$ restricted to $B_{E^{\star}}$.
- (a) Let $f_0 \in B_{E^{\star}}$ and let $V$ be a neighborhood of $f_0$ for $\sigma\left(E^{\star}, E\right)$. We have to find some $r>0$ such that $$ U=\left\{f \in B_{E^{\star}} ; d\left(f, f_0\right)<r\right\} \subset V. $$
As usual, we may assume that $V$ has the form $$ V=\left\{f \in B_{E^{\star}} ;|\langle f-f_0, y_i\rangle|<\varepsilon \quad \forall i=1,2, \ldots, k\right\} $$
with $\varepsilon>0$ and $y_1, y_2, \ldots, y_k \in E$. Without loss of generality we may assume that $\left\|y_i\right\| \leq 1$ for every $i=1,2, \ldots, k$. For every $i$ there is some integer $n_i$ such that $$ \left\|y_i-x_{n_i}\right\|<\varepsilon / 4 $$
(since the set $\left(x_n\right)_{n \geq 1}$ is dense in $\left.B_E\right)$. Choose $r>0$ small enough that $$ 2^{n_i} r<\varepsilon / 2 \quad \forall i=1,2, \ldots, k . $$
We claim that for such $r, U \subset V$. Indeed, if $d\left(f, f_0\right)<r$, we have $$ \frac{1}{2^{n_i}}\left|\langle f-f_0, x_{n_i}\rangle\right|<r \quad \forall i=1,2, \ldots, k $$
and therefore, $\forall i=1,2, \ldots, k$, $$ \left|\langle f-f_0, y_i\rangle\right|=\left|\langle \color{blue}{f-f_0}, y_i-x_{n_i}\rangle+\langle f-f_0, x_{n_i}\rangle\right|<\color{blue}{\frac{\varepsilon}{2}}+\frac{\varepsilon}{2} . \qquad \color{blue}{(\star)} $$
It follows that $f \in V$.
- (b) Let $f_0 \in B_{E^{\star}}$. Given $r>0$, we have to find some neighborhood $V$ of $f_0$ for $\sigma\left(E^{\star}, E\right)$ such that $$ V \subset U=\left\{f \in B_{E^{\star}} ; d\left(f, f_0\right)<r\right\} . $$
We shall choose $V$ to be $$ V=\left\{f \in B_{E^{\star}} ; |\langle f-f_0, x_i\rangle|<\varepsilon \quad \forall i=1,2, \ldots, k\right\} $$ with $\varepsilon$ and $k$ to be determined in such a way that $V \subset U$. For $f \in V$ we have $$ \begin{aligned} d\left(f, f_0\right) &=\sum_{n=1}^k \frac{1}{2^n}\left|\langle f-f_0, x_n\rangle\right|+\sum_{n=k+1}^{\infty} \frac{1}{2^n}\left|\langle \color{blue}{f-f_0}, x_n\rangle\right| \\ &<\varepsilon+\color{blue}{2} \sum_{n=k+1}^{\infty} \frac{1}{2^n}=\varepsilon+\frac{1}{2^{k-1}} \quad \color{blue}{(\star \star)}. \end{aligned} $$
Thus, it suffices to take $\varepsilon=\frac{r}{2}$ and $k$ large enough that $\frac{1}{2^{k-1}}<\frac{r}{2}$.
Let $F$ be a non-empty bounded (in norm) subset of $E^*$. WLOG, we assume $F$ is bounded by $\alpha>0$. We consider the map $$ \varphi:F \to B_{E^*}, f \mapsto \frac{f}{\alpha}. $$
Notice that $E^*$ together with its linear structure and the weak$^*$ topology $\sigma(E^*, E)$ is a locally convex Hausdorff topological vector space. This implies the map $\varphi$ is a homeomorphism (w.r.t. $\sigma(E^*, E)$) from $F$ onto its image $\varphi(F) \subset B_{E^*}$. Clearly, $\varphi(F)$ and thus $F$ is metrizable in $\sigma(E^*, E)$. This completes the proof.