What is the topological dual of a dual space with the weak* topology?

Let $E$ be a locally convex space with topological dual $E'$. Equip $E'$ with the weak*-topology.

Every continuous linear functional $\varphi \colon E' \to \mathbb C$ is of the form $f \mapsto f(e)$ for some $e \in E$, so $(E',\text{weak*})' = E$. In particular, the dual space of $M(X)$ with the weak*-topology is $C(X)$ again.

Since $\varphi$ is continuous, the set $U = \left\{f \in E' \mid \lvert \varphi(f) \rvert \lt 1\right\}$ is an open neighborhood of $0$. Thus, recalling how the basis of neighborhoods at a point in $(E',\text{weak*})$ is defined, there exist $e_1,\dots,e_n \in E$ and $\varepsilon \gt 0$ such that $V = \left\{f \in E' \mid \max\left\{\lvert f(e_i)\rvert \mid i = 1,\dots,n\right\} \lt \varepsilon\right\}$ is contained in $U$. Let $\varphi_i(f) = f(e_i)$.

From $V \subseteq U$ it follows that $\ker{\varphi} \supseteq \bigcap_{i=1}^n \ker{\varphi_i}$ and from linear algebra (see here) we deduce that $\varphi = \sum_{i=1}^n \lambda_i \varphi_i$ for some $\lambda_i \in \mathbb{C}$. In other words, $$ \varphi(f) = \sum_{i=1}^n \lambda_i \varphi_i(f) = \sum_{i=1}^n \lambda_i f(e_i)= f\left(\sum_{i=1}^n \lambda_i e_i \right)$$ and we've shown that $\varphi$ is evaluation at $e = \lambda_1 e_1 + \cdots + \lambda_n e_n \in E$.