In proving that weak topology is not metrizable, I come across below lemma. Could you have a check on my attempt?
Let $(E, | \cdot|)$ be an infinite-dimensional normed linear space and $E^\star$ its continuous dual space. If $X \subseteq E$ is nonempty and open in the weak topology, then $X$ is unbounded in norm.
My proof: First, there exists $a \in X$. Then $X$ is a neighborhood of $a$ in the weak topology. By construction of the weak topology, there are $f_1, \ldots, f_n \in E^\star$ and $\varepsilon_1, \ldots, \varepsilon_n >0$ such that $$X \supseteq U:=\{x\in X \mid \forall k = 1, \ldots,n: |\langle f_k, x-a \rangle| < \varepsilon_k\}.$$
Second, let's prove that there is $0\neq b\in E$ such that $b \in \bigcap_{k=1}^n \ker f_k$. Consider the continuous linear map $$T:E \to \mathbb R^n, x \mapsto \big ( \langle f_1, x \rangle, \ldots, \langle f_n, x \rangle \big ).$$ If $\bigcap_{k=1}^n \ker f_k = \ker T = \{0\}$, then $T$ is injective and thus $\dim T \le n$, which is a contradiction. It follows that $x_n := a+nb \in U$ for all $n\in \mathbb N$. Clearly, $\lim_n |x_n| = \infty$. This completes the proof.