The weak topology of a Banach space $X$ is the locally convex topology associated to the family of semi-norms $$ p_f(x)= |f(x)|, \qquad f\in X^*. $$ If $X^*$ is separable, it obviously suffices to take a countable family of semi-norms associated to a dense sequence in $X^*$. I have found the following two statements which I cannot reconcile:
A locally convex vector space is metrisable if and only if there exists a countable family of semi-norms which induces the locally convex topology.
The weak topology on $X^*$ is never metrisable. (It is metrisable on norm-bounded sets if and only if $X^*$ is separable.)
Now the first result indicates that the weak topology should be metrisable when $X^*$ is separable, but the second one says it is not. Both proofs use a Frechet-type construction $$ d(x,y) = \sum_{k=1}^\infty 2^{-k} \frac{p_k(x-y)}{1 + p_k(x-y)}\quad \text{or}\quad d(x,y) = \sum_{k=1}^\infty 2^{-k}\,|f_k(x-y)| $$ respectively. I can see that in a space with the first kind of metric, there is always a sequence which is unbounded in the sense of topological vector spaces, but which converges to zero with respect to the metric. This cannot happen for weakly convergent sequences due to the Banach Steinhaus theorem.
I have also seen and understood the proof that the weak topology does not admit any kind of metric, but I do not see how the general theorem about the metrisability of locally convex spaces fails here.