See the following post.
I'm struggling to show that $\beta(B)\subseteq C(X)$ is closed.
Here $B$ is a Banach space, $X$ is the closed unit ball in the dual space $B^{*}$ with the weak-star topology and $\beta:B \rightarrow C(X)$ such that $\beta(f)(\phi) = \phi (f)$ for each $\phi\in X$ and each $f \in B$.
I know that $\beta$ is linear, injective, and an isometry (and therefore also continuous/bounded). Also, $C(X)$ is a Banach space with the uniform norm.
There is a result that states that if $\beta(B)$ is a finite-codimensional subspace of $C(X)$, then $\beta(B)$ is closed. Not sure if it's applicable here.
Suppose $\lim_{n\to\infty}\beta(f_n)=\Gamma\in C(X)$. We want to show $\Gamma$ is in the image of $\beta$.
Suppose the $(f_n)_n$ converge to some $f\in B$, in either the strong or weak topologies. Then easily $\Gamma=\beta(f)$ would hold. Fix $\epsilon>0$ and suppose $N\in\Bbb N$ is so large that $\|\beta(f_n)-\beta(f_m)\|_\infty<\epsilon$ on $C(X)$ for $n,m\ge N$.
By Hahn-Banach, there is a functional $\phi$ with norm one such that $\phi(f_n-f_m)=\|f_n-f_m\|$, assuming $f_n\neq f_M$. But then we know $\|f_n-f_m\|=|\beta(f_n)(\phi)-\beta(f_m)(\phi)|<\epsilon$. It follows the sequence is Cauchy and indeed there is $f\in B$, $\lim_{n\to\infty}f_n=f$.