It can be found here a proof of Stone-Weierstrass Theorem through Hahn-Banach theorem (hyperplane separation of convex sets). I find one line in the proof difficult to understand:
Suppose, for the sake of reaching a contradiction, that $(A^\perp)_1 \neq \{0\}$. Clearly, $(A^\perp)_1$ is a compact convex subset of $M(X)_1$. By the Krein–Milman Theorem, $(A^\perp)_1$ has an extreme point $\nu$.
Why must $(A^\perp)_1$ be compact? It is certainly convex and closed, but there does not seem to be any reason why it should be finite dimensional (it is compact only if it is finite dimensional).
It is not compact in the norm topology. The proof is using weak* topology and$(A^{\perp})_1$ is compact in this topology by Banach Alaoglu Theoem.