$\hspace{0.44cm}$As the title says, I am trying to show that given a compact Hausdorff topological space $(S, \tau)$, if $C(S)$ is weakly complete then $S$ is a finite set (i.e. $C(S)$ is finite-dimensional). Let $C(S)_{\leq 1}$ be the norm closed unit ball and $\alpha$ be the canonical mapping from $C(S)$ to $C(S)^{\ast\ast}$.
$\hspace{0.44cm}$My attempt is by using contradiction. Assume $C(S)$ is weakly complete and $C(S)$ is infinite-dimensional. Hence given a fixed $\epsilon > 0$ there will be a countable set $\{f_n\}_{n \geq 1}$ where $\|f_n - f_m\| > \epsilon\,\forall\,n \neq m$. Since weakly complete implies weakly close, according to Goldstine Theorem, and the fact that convergence in $C(S)_{\leq 1}$ in weak topology is equivalent to convergence in $\alpha[C(S)_{\leq 1}]$ in weak-* topology, we then can conclude $C(S)$ is reflexive and $f_n$ converge to $g \in C(S)$ pointwise. Then I fail to obtain a contradiction here.
$\hspace{0.44cm}$I plan to use the Arzelà–Ascoli theorem but then fail to show that $\{f_n\}_{n \geq 1}$ is equi-continuous. Also it is not obvious that $f_n$ converge to $g$ in norm. Any hints will be appreciated.
