Metrizability of the weak topology $\sigma(E,E')$ of a normed space $E$ Let $(E , \left \| \cdot \right \|_E)$ a normed space with topology $\mathcal{T}_E$ induced by norm. We have that
(1) If $\mathrm{dim}(E) < \infty$ then  $\sigma(E,E') = \mathcal{T}_E$ and weak topology $\sigma(E,E')$ is metrizable.
(2) If $\mathrm{dim}(E)=\infty$ then $\sigma(E,E') \neq  \mathcal{T}_E$ and weak topology $\sigma(E,E')$ is not metrizable.
I have some doubts on the demonstration of the point (2)
$Proof$ $(2)$.
Let $\mathrm{dim}(E)=\infty$, then it is sufficient to prove that each $\sigma(E,E')$-open neighborhood if the origin contains a infinite-dimensional subspace, consequently the weak topology can not be induced by any norm (why it can not be induced by any norm?). If
$u_i \in E'$ $\forall i=1,...,n$ and $\epsilon > 0$, $\sigma(E,E')$-open neighborhood $U_n$ has the form
\begin{align*}
\displaystyle U_n=\lbrace x \in E : \max\lbrace p_{u_1}(x),...,p_{u_n}(x) \rbrace < \epsilon \rbrace = \lbrace x \in E : |u_i(x)| < \epsilon  \rbrace= \bigcap_{i=1}^n B_{\mathbb{K}}^{u_i}(\epsilon)
\end{align*}
$\forall x \in E$, let $u(x):=(u_1(x),...,u_n(x))$ a homeomorphism with values in $\mathbb{K}^n$, from dimensional equation we have that $\mathrm{dim}(E)=\mathrm{dim}N(u) + \mathrm{dim} R(u)$, but $\mathrm{dim}R(u) \leq n$ and then $\mathrm{dim}N(u)=\infty$ necessarily, and also $N(u) \subset U_n$.

It boils down to this: 
Metrizable topological vector spaces (and hence normed spaces) are locally bounded, and what this argument shows is that the $\sigma(E,E')$ topology on $E$ is not locally bounded.  
To see this, pick $x_0\in N(u)$ with $x_0\neq0$.  The Hahn-Banach theorem furnishes some $v\in E'$ with $v(x_0)=\|x_0\|_E$ and $|v(x)|\leq\|x\|_E$ for all $x\in E$.  Put
$$ V=\{x\in E:|v(x)|<1\} $$
Then $V$ is a $\sigma(E,E')$ neighborhood of $0$, and $U_n\not\subset tV$ for any $t>0$.  Since every open neighborhood of $0$ contains an open neighborhood of the form $U_n$, the $\sigma(E,E')$ topology is not locally bounded.
