It is known that if $X$ is a separable Banach space with dual $X^\ast$, then $B_{X^\ast}$, the closed the unit ball in $X^\ast$, is compact and metrizable in the weak* or $\sigma(X^\ast, X)$-topology. In particular, $B_{X^\ast}$ is weak* separable and thus $X^\ast$ is also weak* separable.
If $X$ is only a separable Fréchet space, is it true that its dual $X^\ast$ is $\sigma(X^*,X)$-separable?
Let $X$ be a metrizable and separable LCTVS. It is topologized by a countable family of seminorms $\{p_i\}_{i\in\mathbb N}$, and it has a countable dense subset $\{x_j\}_{j\in\mathbb N}$. By Hahn-Banach, for each $(i,j)\in\mathbb N^2$ there exists a functional $f_{ij}\in X^*$ such that $\langle f_{ij},x_j\rangle=p_i(x_j)$ and $\langle f_{ij},x\rangle\le p_i(x)$ for all $x\in X$. Consider $Y:=\text{span}\{f_{ij}:(i,j)\in\mathbb N^2\}\subset X^*$: I claim that for every $x\in X$, if $\langle f_{ij},x\rangle=0$ for all $(i,j)\in\mathbb N^2$, then $x=0$. By the usual criterion, this is equivalent to $Y$ being $\sigma(X^*,X)$-dense (one has $\overline Y^{\sigma(X^*,X)}=(Y_\perp)^\perp$, that is the whole $X^*$ iff $Y_\perp=(0)$, which is the claim).
Proof of the claim: Assume $x\in X$ has $\langle f_{ij},x\rangle=0$ for all $(i,j)\in\mathbb N^2$. For every $i\in\mathbb N$ we have $$p_i(x_j)=\langle f_{ij},x_j\rangle= \langle f_{ij},x_j-x\rangle\le p_i(x_j-x).$$ Since $\{x_j\}_{j\in\mathbb N}$ is dense in $X$, $x_j\to x$ along a subsequence, which implies $p_i(x)=0$ by the above inequality, so also $x=0$.
