Maybe I am oversimplifying things, but to me the proof looks straight forward.
The function $f$ mapping a policy $\pi$ to its value function $V^\pi$ is surjective. It is therefore enough to show that the optimal value function $V^*$ exists. The optimal policy then exists, too, because $f^{-1}\{V^*\}\neq\emptyset$.
The value function has the form $V:S\rightarrow\mathbb{R}$ where $S$ is the finite set of states. A finite, discrete set is compact. Further, we can define the isolated points metric on $S$, i.e.
\begin{equation}
d_S(x,y):=\begin{cases}
1 &, y\neq x \\
0 &, y=x
\end{cases}
\end{equation}
If $S$ is a metric space, we can show that $V$ is continuous [1]. The idea here is that if the sequence $s_n$ in $S$ converges to $s\in S$ we can choose $\epsilon \in (0,1)$ and get $s_n=s~~\forall~n>N_\epsilon$ (which does exist).
A continuous function $V$ on a compact metric space $S$ attains its maximum at some point in $S$ [1]. Hence $V^*$ exists. $\Box$
[1] https://www.rose-hulman.edu/~bryan/lottamath/compact.pdf
Arguably this gets more interesting if $S$ is no longer finite. There we have no guarantee that $V$ will be continuous. I assume you can still show the existence of $pi^*$ if the reward is bounded, but I've never written it down or thought it through thoroughly.