Rudin Functional Analysis Theorem 3.25

This is probably the first question regarding the theorem in the title...

If $$K$$ is a compact set in a locally convex space $$X$$, and if $$\bar{co}(K)$$ is also compact, then every extreme point of $$\bar{co}(K)$$ lies in $$K$$.

I struggle to understand the very first 3 lines of the proof.

Assume that some extreme point $$p$$ of $$\bar{co}(K)$$ is not in $$K$$. Then there is a convex balanced neighborhood $$V$$ of $$0$$ in $$X$$ such that $$(p + \overline{V}) \cap K = \emptyset$$

Can anyone explain why can we find such a $$V$$? My guess is that it's due to compacteness of both $$\overline{co}(K)$$ and $$K$$ plus the fact that $$K \subset \overline{co}(K)$$.

Compactness is not required. $$K^{c}$$ is an open set containing $$p$$ and this gives existence of such a set $$V$$. This has been proved in earlier sections of Rudin's book$. • The subtlety I'm not getting is the closure of$V$not$V$itself. Feb 20 '19 at 10:34 •$V$is an open neighborhood of$0$. It is fact that every open neighborhood of$0$contains the closure of another open neighborhood of$0\$. This is what is being used here. Feb 20 '19 at 10:37