Dual space $E^*$ metrizable iff E has a countable basis 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.
 A: If I’m not mistaken, you’ve essentially identified $E^*$ with $^I\mathbb{R}$, the family of all real-valued functions on the index set $I$, equipped with the seminorms $p_i(f)=|f(i)|$ for $i\in I$. For each finite $F\subseteq I$, $\epsilon>0$, and $f\in ^I\mathbb{R}$ let $$\begin{align*}B_{F,\epsilon}(f)&=\{g\in^I\mathbb{R}:\forall i\in F[p_i(g-f)<\epsilon]\}\\
&=\{g\in^I\mathbb{R}:\forall i\in F[|g(i)-f(i)|<\epsilon]\};
\end{align*}$$ the family of all such $B_{F,\epsilon}(f)$ is a local base at $f$. 
Recall that every metric space is first countable, meaning that each point has a countable local base. Suppose that $\{V_k:k\in\omega\}$ is a local base at $f\in^I\mathbb{R}$. Then for each $k\in\omega$ there must be a finite $F_k\subseteq I$ and an $n_k\in\omega$ such that $B_{F_k,2^{-n_k}}(f)\subseteq V_k$. It follows that $$\{f\}\subseteq\bigcap_{k\in\omega}B_{F_k,2^{-n_k}}(f)\subseteq\bigcap_{k\in\omega}V_k=\{f\}$$ and hence that $$\bigcap_{k\in\omega}B_{F_k,2^{-n_k}}(f)=\{f\}\;.$$
Let $C=\bigcup\limits_{k\in\omega}F_k$; clearly $C$ is countable, so we can choose $i_0\in I\setminus C$. Now define $g\in^I\mathbb{R}$ by $$g(i)=\begin{cases}
f(i),&i\ne i_0\\
f(i_0)+1,&i=i_0\;;
\end{cases}$$ clearly $g\ne f$, and it’s not hard to see that $$g\in\bigcap_{k\in\omega}B_{F_k,2^{-n_k}}(f)\;.$$
This is the desired contradiction.
By the way, $^I\mathbb{R}$ with this topology is easily seen to be homeomorphic to the product topology on $\mathbb{R}^{|I|}$, so the problem really boils down to showing that $\mathbb{R}^\kappa$ is metrizable iff $\kappa\le\omega$.
