Let $E$ be a Banach space, $E^*$ its dual space, and $w^*$ the weak$^*$ topology of $E^*$. For a subset $G$ of $E^*$, let $w^*_G$ be the subspace topology of $G$ that is induced from $w^*$. A subset $G$ of $E^*$ is said to be pseudo-absorbing if for each $f \in E^*$ there exists $r \neq 0$ such that $rf \in G$. Clearly, the open/closed unit ball is pseudo-absorbing.
I'm trying to generalize Theorem 3.28. in Brezis's Functional Analysis, i.e.,
Theorem: Let $G \subset E^*$ such that $0 \in G$. If $G$ is pseudo-absorbing and $w^*_G$ is metrizable, then $E$ is separable.
Could you have a check on my attempt?
Proof: Assume $w^*_G$ is metrizable by a metric $d$ on $G$. Let $$ U_n := \left \{ f \in G ; d(f, 0) < \frac{1}{n+1} \right \} \quad \forall n \in \mathbb N. $$
Then $U_n$ is a neighborhood (nbh) of $0$ in $w^*_G$. Then there is a nbh $W_n$ of $0$ in $w^*$ such that $U_n = W_n \cap G$. Then there is a nbh $V_n$ of $0$ in $w^*$, with the form $$ V_n = \{f \in E^* ; |\langle f, x\rangle| < 1 \text{ for all } x \in D_n\}, $$ for some finite subset $D_n$ of $E$, such that $V_n \subset W_n$. Let $K_n := V_n \cap G$. Then $K_n \subset U_n$ and $K_n$ is a nbh of $0$ in $w^*_G$. Clearly, $$ K_n = \{f \in G ; |\langle f, x\rangle| < 1 \text{ for all } x \in D_n\}. $$
Let $D := \bigcup_n D_n$. Then $D$ is countable. Let $F$ be a vector space spanned by $D$. Then $F$ is separable. It suffices to prove that $F$ is dense in $E$. Fix $f\in E^*$ such that $f(x)=0$ for all $x \in D$. By Hahn-Banach theorem, it suffices to prove $f=0$. Because $G$ is pseudo-absorbing, there is $r \neq 0$ such that $rf \in G$. It follows that $rf \in K_n$ and thus $rf\in U_n$ for all $n$. So $rf=0$ and thus $f=0$. This completes the proof.