I'm trying to prove the following proposition:
Let $\left(f_{n}\right)$ be a sequence in $E^{\star}$ such that for every $x \in E,\left\langle f_{n}, x\right\rangle$ converges to a limit. Prove that there exists some $f \in E^{\star}$ such that $f_{n} \stackrel{\star}{\rightharpoonup} f$ in $\sigma\left(E^{\star}, E\right)$ (weak* convergence).
First construct $f:E\to \mathbb{R}$ such that $x\mapsto \lim\limits_{n\to \infty}\langle f_n,x \rangle$. Therefore $\langle f,x \rangle=\lim\limits_{n\to \infty}\langle f_n,x \rangle$, thus we only (only?) need to prove $f\in E^*$ by the definition of weak* convergence.
Since $f_n(x)$ converges $\forall x\in E$, we have $$\sup_n|f_n(x)|<\infty,\ \forall x\in E.$$
From uniform boundedness theorem we have $c:=\sup\limits_n||f_n||<\infty$, which means $|f_n(x)|\leq c||x||,\ \forall x\in E,\ \forall n$. Take $n\to \infty$ we get $|f(x)|\leq c||x||,\ \forall x\in E$, which is desired.
Is my proof reasonable?
If $E$ is complete your proof is fine. Otherwise the result is not true. Here is a counter-example: let $E$ be the space sequences of real numbers with at most a finite number of non -zero terms. Consider the norm from $\ell^{2}$ on this and define $f(x)=(x_1,\frac {x_2} {\sqrt 2}, \frac {x_3} {\sqrt 3},...)$. Then $f_n(x)$ converges for each $x$ but continuity of the limit $f$ would imply that the sequence $(1,\frac 1 {\sqrt 2}, \frac 1 {\sqrt 3},...) \in \ell^{2}$ which is not true. [I have used the fact that ant continuous linear functional on $E$ extends to a continuous linear functional on $\ell^{2}$).