Here is Greg's proof of the Krein-Milman theorem (only the existence part). My apologies about the TeX: i don't know what does or doesn't work on this site, so I've excluded the begin…end environments, but placed headings to guide the reader.
Proposition 1: Let $A\subseteq V$ be a non-empty compact convex set of an hausdorff locally convex semi-topological vector space (over some field which contains the reals topologically). Then $A$ has an extreme point.
Specifically we shall show that every non-empty convex open-in-$A$ proper subset $U^{-}$, has an extension to a convex open-in-$A$ proper subset $U^{+}\supseteq U^{-}$, for which $A\setminus U^{+}={e}$ some extreme point $e$.
Proof (Prop 1): If $A$ is a singleton, we are done. So suppose otherwise. First notice we can separate two points by a convex open set, $U$, yielding $U^{-}=U\cap A$ a convex proper subset open in $A$. So fix such a set $U^{-}$.
Now let $\mathbb{P}:=\{W:U^{-}\subseteq W\mbox{ convex proper subset open in }A\}$.
We claim that the p.o. $(\mathbb{P},\subseteq)$ has the Zorn property.
Clearly the union of any chain of convex subsets open in $A$,
is again a convex subset open in $A$. Suppose \it{per contra} that
a union of a chain $\mathcal{C}\subseteq\mathbb{P}$ is equal to $A$.
Then since $A$ is compact, there must be a finite sub(chain) $\mathcal{C}'\subseteq\mathcal{C}$
for which $\operatorname{max}(\mathcal{C}')=\bigcup\mathcal{C}'=A$, which contracts
the properties of members of $\mathbb{P}$. Thus $\mathbb{P}$ is non-empty
and has the Zorn property. So fix $U^{+}$ open such that $U^{+}\cap A\in\mathbb{P}$ maximal.
\textbf{Sub-claim 1:}
$U^{+}\cap A\subset\operatorname{cl}_{A}U^{+}$ strictly.\\
\textbf{Proof (sub-claim 1):}
Let $x\in U^{+}\cap A$ and $y\in A\setminus U^{+}$.
Consider the map (here we need the underlying field to contain the reals topologically)\\
\begin{math}\begin{array}[t]{lrclll}
&f &: &\mathbb{F} &\to &V\\
&&: &s &\mapsto &s~x+(1-s)~y\\
\end{array}\end{math}\\
which is continuous and affine.
Thus the set $S:=\{s\in[0,1]:s\mapsto s~x+(1-s)~y\in U^{+}\cap A\}
=[0,1]\cap f^{-1}U^{+}\cap f^{-1}A)$
is a convex. The set $f^{-1}A$ is closed convex and contains $0,1$,
thus contains $[0,1]$. So $S=[0,1]\cap f^{-1}U^{+}$ which is convex open
in $[0,1]$. Since $0\notin S$ and $1\in S$, then $S=(a,1]$ some $a\in[0,1)$.
Clearly $f(a)=lim_{s\searrow a}f(s)$ which is a limit of vectors from $U^{+}\cap A$,
and thus lies in $\operatorname{cl}U^{+}$. It also clearly lies in $A$,
since $f^{-1}A\supseteq[0,1]$. We also have $f(a)\notin U^{+}$.
Thus $\operatorname{cl}_{A}U^{+}\setminus U^{+}
=A\cap\operatorname{cl}U^{+}\setminus U^{+}
\supseteq\{f(a)\}$.
Thus the containment is strict.
QED (sub-claim 1)
\textbf{Sub-claim 2:}
If $W\subseteq A$ is convex, then $U^{+}\cup W$ is convex.\\
\textbf{Proof (sub-claim 2):}
Fix $x\in U^{+}\cap A,t\in(0,1)$. Consider the map\\
\begin{math}\begin{array}[t]{lrclll}
&T &: &V &\to &V\\
&&: &y &\mapsto &t~x+(1-t)~y\\
\end{array}\end{math}\\
This is continuous and affine. By the proposition below,
we also see that $T(\operatorname{cl}(U^{+}))\subseteq U^{+}$.
So $A\cap T^{-1}U^{+}$ is convex, open in $A$, and contains $A\cap\operatorname{cl}(U^{+})$
which strictly contains $U^{+}\cap A$ by Claim 1. Thus by maximality of $U^{+}$ in $\mathbb{P}$,
$A\cap T^{-1}U^{+}\overset{\mbox{\tiny must}}{=}A$. In particular, $T^{-1}U^{+}\supseteq A$,
and so $TW\subseteq TA\subseteq U^{+}$. Utilising such maps as $T$,
we see that a convex-linear combination of any pair of
elements from $U^{+}\cup W$ is contained in $U^{+}\cup W$.
QED (sub-claim 2)
\textbf{Sub-claim 3:}
$A\setminus U^{+}$ is a singleton.\\
\textbf{Proof (sub-claim 3):}
Else, let $x_{1},x_{2}\in A\setminus U^{+}$ distinct.
Let $W$ be a convex open set separating $x_{1}$ from $x_{2}$.
Then by Claim 3, $U^{+}\cap A\cup W\cap A$ is convex open in $A$.
Since $x_{1}\in W\setminus U^{+}$, this convex set is strictly large than
$U^{+}\cap A$. By maximality of $U^{+}$ in $\mathbb{P}$,
we have that $U^{+}\cap A\cup W\cap A=A\ni x_{2}$. But this contradicts
the fact that $x_{2}\notin U^{+}\cup W$.
QED (sub-claim 3)
At last, we claim that the point $e\in A\setminus U^{+}$ is extreme in A.
Consider $x,y\in A$
and $t\in(0,1)$ for which $tx+(1-t)y=e$. Case by case, we see
that if $x,y\in U^{+}$ then $e=tx+ty\in U^{+}$.
If $x\in A\setminus U^{+}$, $y=\frac{e-tx}{1-t}=\frac{e-te}{1-t}=e=x$.
If $y\in A\setminus U^{+}$, then $x=e=y$ similarly.
Thus $e$ is extreme.
QED(Prop 1)
Proposition 2: Let $A\subseteq V$ be a convex subset of a semi-topological space, and $x,y$ be vectors lying respectively in the closure and interior of $A$. Then for any $t\in(0,1)$, we have $tx+(1-t)y\in A$.
Proof: Let $U$ be an open neighbourhood of $y$, contained in $A$. Observe that $x-t^{-1}(1-t)(U-y)$ is an open neighbourhood of $x$, and thus meets $A$, say at $x'$. Thus $x\in x'+t^{-1}(1-t)(U-y)$ implying $tx+(1-t)y\in tx'+(1-t)U \subseteq tA+(1-t)A \subseteq A$, since $A$ is convex. QED(Prop 2)
