Proof of Stone Weierstrass Theorem from Hahn Banach

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).