Why the weak topology is never metrizable?

Note that we require $X$ to be infinite-dimensional. For if $X$ is finite-dimensional, there is only one vector topology on $X$, which is metrizable.

Q1: That's correct. If the metric $d$ induces the weak topology, then the balls in that metric are weakly open.

Q2: This is part of a more general result that every weakly open subset of $X$ is unbounded in the norm-topology of $X$. The idea is that if $U$ is weakly open and $x_0\in U$, then $U$ there is some $\varepsilon>0$ and $\varphi_1,\ldots,\varphi_n\in X^*$ such that $$x_0\in\bigcap_{k=1}^n\{x\in X:|\varphi_k(x-x_0)|<\varepsilon\}\subset U.$$ and the set $\bigcap_{k=1}^n\{x\in X:|\varphi_k(x-x_0)|<\varepsilon\}$ contains the hyperplane $x_0+\bigcap_{k=1}^n\ker(\varphi_k)$, which is unbounded in the norm-topology of $X$.