# The Weak topology on an infinite-dimensional space is not metrizable

Let $X$ be an infinite-dimensional normed space. I want to prove that the weak topology on $X$ is not metrizable. This is my solution:

Assume that there is a metric $d$ on $X$ inducing the weak topology, and consider $U_n:=\{x\in X: d(x,0)<\frac{1}{n}\}$. We know each $U_n$ is weakly open and so will be unbounded (for the strong topology), and thus $$\forall (n)\exists (x_n\in U_n) \:\text{s.t.}\: \|x_n\|\geq n$$ But $x_n\to 0$ in $(X, d)$, so that $x_n \stackrel{w}{\to}0$, and hence $(x_n)$ is bounded. Contradiction.

Is this the right solution?