I'm having trouble even figuring out how to get my hands on this problem. So the statement is:
Let $K$ be a compact metric space. Prove: $C(K)$ is reflexive $\iff K$ is a finite set. [Suggestion: it may be helpful to show (and use) that for each $x\in K$ the set function $\delta_x$ on $\operatorname{Bor}(K)$ be defined by $$\delta_x(A)=\begin{cases}1&\text{if }x\in A\\0&\text{if }x\in K\backslash A\end{cases}$$ is a regular measure on $\operatorname{Bor}(K)$.]
I understand this is going to be an application of the Riesz Representation Theorem but I just can't seem to put it together.
Every bounded Borel function $f: K\to\mathbb C$ gives an element of $C(K)^{**}=M(K)^*$ via $\mu\mapsto \int f\, d\mu$. If $C(K)$ is reflexive, then there must be a continuous $g$ such that $\int f\, d\mu = \int g\, d\mu$ for all $\mu\in M(K)$. By testing this on $\mu=\delta_x$ (the hint in action here, though it seems we would have done this anyway), we conclude that $f(x)=g(x)$. In other words, every bounded Borel function on $K$ is continuous. Since we can take $f=\chi_{a}$, this implies that every subset of $K$ is open. Since $K$ is also compact, it must be finite.
The converse is clear because $K$ finite implies that $C(K)$ is finite dimensional.