$\text{dim}\;E<\infty$ if and only if $E'$, in the weak topology, is normable Let $E$ be a locally convex Hausdorff space over $\mathbb{C}$. I want to prove that: $\text{dim} \;E<\infty$ if and only $E'_{\sigma}$ is normable, where $E'_{\sigma}$ denotes the dual space of $E$ with the weak topology.
The part " $\text{dim} \;E< \infty$ implies $E'_{\sigma}$  normable" I managed to prove.
But the converse I couldn't prove. The only idea I had to prove this claim was to use Riesz’s theorem (see here Theorem $6.5$, page $160$).
However I haven't been able to progress, because to start I can not prove that $ E $ is normed space.
 A: 
However I haven't been able to progress, because to start I can not prove that $E$ is normed space.

And to prove that, you must prove that $E$ is finite-dimensional, because otherwise it might not be normable.
Thus we need a different idea. An important fact that is quite useful here is that in a weak topology (that is, any topology $\sigma(E,F)$ obtained from a bilinear pairing $E \times F \to \mathbb{C}$) every neighbourhood of $0$ contains a linear subspace of finite codimension. By definition, every $\sigma(E,F)$-neighbourhood of $0$ contains a set of the form
$$V(f_1, \dotsc, f_n) := \{ e \in E : \lvert f_{\nu}(e)\rvert < 1 \text{ for } 1 \leqslant \nu \leqslant n\},$$
where $f_1, \dotsc, f_n \in F$ (and we view $F$ as a subset of the algebraic dual of $E$ via the pairing to write $f(e)$ rather than $\langle e, f\rangle$, where $\langle\,\cdot\,,\,\cdot\,\rangle \colon E\times F \to \mathbb{C}$ is the pairing). Typically the neighbourhoods are described by $\lvert f_{\nu}(e)\rvert < \varepsilon_{\nu}$, but replacing $f_{\nu}$ with a suitable multiple of it we can without loss of generality assume $\varepsilon_{\nu} = 1$ for all $\nu$. Considering the map
$$\Phi \colon e \mapsto \bigl(f_1(e), \dotsc, f_n(e)\bigr)$$
from $E$ to $\mathbb{C}^n$ it is clear that $\ker \Phi \subseteq V(f_1,\dotsc, f_n)$, and since
$$E/\ker \Phi \cong \operatorname{im} \Phi \subseteq \mathbb{C}^n\,,$$
the linear subspace $\ker \Phi$ has finite codimension ($\leqslant n$) in $E$.
On the other hand, if $(X, \lVert \,\cdot\,\rVert)$ is a normed space, a ball
$$B_r(0) := \{ x \in X : \lVert x\rVert < r\}$$
with finite radius $r > 0$ contains no nontrivial linear subspace. For if $x \neq 0$, then $\lVert x\rVert > 0$, and $z\cdot x \notin B_r(0)$ for all $z$ such that $\lvert z\rvert \geqslant r/\lVert x\rVert$.
If $E'_{\sigma}$ is normable, then there are $f_1, \dotsc, f_n \in E'' = (E'_{\sigma})'$ such that
$$V(f_1,\dotsc, f_n) \subseteq B_1(0)\,. \tag{$\ast$}$$
Now $V(f_1, \dotsc, f_n)$ contains a linear subspace of finite codimension. But the only linear subspace contained in $B_1(0)$, and a fortiori the only linear subspace contained in $V(f_1, \dotsc, f_n)$, is the trivial subspace $\{0\}$.
It follows that $\{0\}$ has finite codimension in $E'$, i.e. $E'$ is finite-dimensional. (And if $E'$ is finite-dimensional, then $E'_{\sigma}$ is normable, since there is only one Hausdorff vector space topology on finite-dimensional spaces [over $\mathbb{C}$ or $\mathbb{R}$].)
Now to conclude that $E$ is finite-dimensional too, we need the assumption that $E$ is Hausdorff locally convex. Let $(\lambda_1, \dotsc, \lambda_d)$ be a basis of $E'$ and consider the linear map
$$\Lambda \colon e \mapsto \bigl(\lambda_1(e), \dotsc, \lambda_d(e)\bigr)\,.$$
By the Hahn-Banach theorem, $\Lambda$ is injective since $E$ is Hausdorff locally convex and the component functionals span $E'$, thus
$$\dim E = \dim \operatorname{im} \Lambda \leqslant \dim \mathbb{C}^d = d$$
follows. (And since $d = \dim E'$ is no larger than the dimension of the algebraic dual $E^{\ast}$ and $\dim E^{\ast} = \dim E$ for finite-dimensional $E$, it follows that $\dim E = \dim E' = d$.)
