Let $V$ be a compact space. Consider on the one hand the space $C(V)$ of continuous functions of $V$ and on the other hand the space $M(V)$ of Radon measures of bounded total variation on $V$. Now consider the weakest topology on $C(V)$ that makes all functionals $f \mapsto \int f d\mu$ for $\mu \in M(V)$ continuous.
Take a measure $\nu \in M(V)$. Is the set $\{f \in C(V) : |\int f d\nu| \leq 1\}$ metrizable in the topology defined above?
The topology is the weak topology on $C(V)$ (since $M(V)$ is its dual space). Recall that Weak topology on an infinite-dimensional normed vector space is not metrizable. Restricting to the set $\{f \in C(V) : |\int f d\nu| \leq 1\}$ does not help. Indeed, even restricting to the hyperplane $\{f \in C(V) : \int f d\nu = 0 \}$ leaves us with an infinite-dimensional normed space; so the weak topology is not metrizable by the aforementioned result.