Proof. Assume WLOG that $K\neq\emp$. Let $K_{0}\suf K$ be a clsoed, extreme subset of $K$ and $\cali{E}(K_{0})\suf \cali{P}(K)$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_{0}$. Then $K_{0}\in\cali{E}(K_{0})\neq\emp$. Since $K$ is comapct, for any chain $\cali{C}\suf \cali{E}(K_{0})$, $\caps{}{A\in\cali{C}}A$ is also nonempty, closed and extreme.

By Zorn’s Lemma, there exists a minimal element $A$ of $\cali{E}(K_{0})$. Let $x,y\in A,\phi\in E^{*}$ and $\al=\SUP{z\in A}\inner{z}{\phi}_{E}$, then $\curl{\phi=\al}\cap K$ is a non-empty, closed and extreme subset of $K$ by Lemma 4.3.3, so $A\cap \curl{\phi=\al}$ is also extreme. By minimality of $A,A\suf \curl{\phi=\al}$. Thus by Hahn-Banach, $A$ consists of exactly one point. In which case, for any $y,z\in E$ with $A\suf (y,z)\suf K$, we have $y=z\in A$ thus there exists an extreme point of $K$ in $K_{0}$.

By Proposition 3.10.4, $K$ is closed. By the reasoning above applied to $K$ in place of $K_{0}$, $K$ admits at least one extreme point.

Let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\cl{\text{Conv}}(C)\sub K$, then by Hahn-Banach, there exists $x\in K$ and $\phi\in E^{*}$ such that $\SUP{y\in \cl{\text{Conv}}(C)}\inner{y}{\phi}_{E}<\inner{x}{\phi}_{E}$. Let $\al=\SUP{z\in K}{z,\phi}_{E}$, then by Lemma 4.3.3, $K\cap \curl{\phi=\al}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K\cap \curl{\phi=\al}$, which contradicts the fact that $\SUP{y\in \cl{\text{Conv}}(C)}\inner{y}{\phi}_{E}<\al$.$\square$