Proof, [Theorem 1.12.5, BS17]. Assume without loss of generality that $K \ne \emptyset$.

Let $K_{0} \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_{0}) \subset 2^{K}$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_{0}$. Since $K_{0} \in \mathcal{E}(K_{0})$, $\mathcal{E}(K_{0}) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C}\subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme.

By Zorn’s lemma, there exists a minimal element $A$ of $\mathcal{E}(K_{0})$. Let $x, y \in A$, $\phi \in E^{*}$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha}\cap K$ is a non-empty, closed, and extreme subset of $K$ by Lemma 11.3.3, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the Hahn-Banach Theorem, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_{0}$.

Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point.

Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the Hahn-Banach Theorem, there exists $x \in K$ and $\phi \in E^{*}$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E}< \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by Lemma 11.3.3, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E}< \alpha$.$\square$