I'm trying to prove that the weak topology of an infinite dimensional normed space is not metrizable. Here is what I have come up with, I'm not sure if it is correct, please correct me if it wrong! Thank you! Here, I used $Z(f)$ to denote the kernel of a linear functional $f$, and $'$ to denote the dual space.
Assume that $X$ is infinite dimensional, let $Y$ be the completion of $X$, so $Y$ is a Banach space, $\tau:X\rightarrow Y$ is a normed preserving linear map with dense image. A net $(x_n)\in X$ converges to $x\in X$ in the weak topology of $X$ iff $f(x_n)\rightarrow f(x)$ for all $f\in X'$ iff $g(\tau(x_n))\rightarrow g(\tau(x))$ for all $g\in X'$ iff $\tau(x_n)\rightarrow\tau(x)$ in the weak topology of $Y$. Therefore, the weak topology of $X$ agrees with the subsapce topology of the range of $\tau$ as a subspace of the weak topology of $Y$. Hence if we can show the weak topology of $Y$ is not metrizable, then the weak topology of $X$ is also not metrizable.
Therefore, we can assume that $X$ is an infinite dimensional Banach space, so its dimension must be uncountable. If $X$ is metrizable under the weak topology, then $X$ is first countable at $0$ under the weak topology. This implies that we can find a neighborhood basis of the weak topology of $X$ at $0$, that consists of a sequence of open sets $(U_n)_1^\infty$ each one is of the form $\bigcap_{i=1}^j f_i^{-1}(B(0,\epsilon))$, where $f_i\in X'$ is nonzero and $\epsilon>0$, $j\in\mathbb{Z}^+$. Let $\mathcal{F}$ be the collection of all such $f_i$'s for for all $U_n$'s. $\mathcal{F}$ is a countable set. Since $X$ is of uncountable dimension, it is but a linear algebraic fact that $\bigcap_{f\in\mathcal{F}}Z(f)$ is nonzero. Hence, we can find a nonzero $v\in\bigcap_{f\in\mathcal{F}}Z(f)$. Using the Hahn-Banach extension theorem, we can find some $f\in X'$ such that $f(v)=1$. Then $v\in U_n$ for each $n\in\mathbb{Z}^+$ but $v\notin f^{-1}(B(0,1/2))$. However, $f^{-1}(B(0,1/2))$ is open in the weak topology of $X$, so $U_n\subset f^{-1}(B(0,1/2))$ for some $n\in\mathbb{Z}^+$. Contradiction!