I have trouble proving the following theorem:
If $E$ is a locally convex, Hausdorff topological vector space, then $E^*$ is metrizable if and only if $E$ has an (at most) countable basis.
I've proved the 'if'-part, but I'm stuck with the 'only if' part. I've tried doing it the following way: suppose $E$ doesn't have a countable basis, so $E = \oplus_{i \in I} \mathbb{R}e_i$ with $|I| > \aleph_0$. The topology on $E^*$ is determined by the seminorms $p_x:E^* \rightarrow \mathbb{R},f \mapsto |f(x)|, x \in E$, but since $x$ is in $E$, this is the same as taking $p_i:E^* \rightarrow \mathbb{R},f \mapsto |f(e_i)|$ for each $i \in I$. Now, if $E^*$ would be metrizable, it would be $A_1$, so there would be a countable basis $V_n, n \in \mathbb{N}$ so that $ \forall K \subset I \text{ finite }, \forall k \in K, \forall \epsilon_k > 0 \exists n: V_n \subset \bigcap_{k \in K} \{p_k < \epsilon_k\}. $ I'm trying to get a contradiction, but I don't know how to do it. Any help would be appreciated.