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False. Take the separable Banach space $E={\ell }^1$, whose continuous dual space is ${\ell }^{\infty }$. We prove that first countability of the weak unit ball would force this dual to be norm separable, producing a contradiction.

Suppose that $B$ had a countable weak neighbourhood base $(U_n)$ at zero. Inside each $U_n$, choose a basic relative weak neighbourhood specified by finitely many continuous linear functionals $F_n$. The union $F={\bigcup }_n{F}_n$ is countable. For any $f\in E^{\prime }$ and $ϵ>0$, some $U_n$ lies in $\{x\in B:|f(x)|<ϵ\}$. On the vector subspacethis implies $\Vert f{|}_{M_n}\Vert \le ϵ$, by scaling vectors into $B$. The Hahn-Banach theorem extends $f{|}_{M_n}$ to $h\in E^{\prime }$ with $\Vert h\Vert \le ϵ$. The functional $f-h$ vanishes on $M_n$ and is therefore a linear combination of $F_n$: it factors through the finite-dimensional coordinate map $x\mapsto (g(x){)}_{g\in F_n}$, and a linear functional on that map's image extends to the finite-dimensional coordinate space. HenceEvery $f\in E^{\prime }$ thus belongs to the norm closure of $\mathrm{span}F$. Taking rational coefficients, or rational real and imaginary parts, gives a countable norm-dense subset of $E^{\prime }$. This proves that weak-ball metrizability requires a norm-separable dual.

For ${\ell }^1$, every bounded sequence $a=(a_j)$ defines $f_a(x)={\sum }_j{a}_j{x}_j$ with $\Vert f_a\Vert =\Vert a{\Vert }_{\infty }$. Conversely, the values $f(e_j)$ of any bounded linear functional determine such a sequence and recover $f$ by density of finitely supported vectors. Thus $E^{\prime }={\ell }^{\infty }$. The uncountable family $\{0,1{\}}^{\mathbb{N}}$ in ${\ell }^{\infty }$ has distance one between distinct members, so this dual is not norm separable: disjoint balls of radius less than $1/2$ would require distinct members of any countable dense set. Since every metrizable space is first countable, the weak unit ball of ${\ell }^1$ is not metrizable.