John B. Conway A Course in Functional Analysis :
(page 141)7.4. The Krein-Milman Theorem. If $K$ is a nonempty compact convex subset of $a$ LCS $\mathscr{X},$ then $\operatorname{ext} K \neq \square$ and $K=\overline{\operatorname{co}}(\operatorname{ext} K)$
PROOF. (Léger $[1968] .)$ Note that $(7.3 \mathrm{e})$ says that a point $a$ is an extreme point if and only if $K \backslash\{a\}$ is a relatively open convex subset. We thus look for a maximal proper relatively open convex subset of $K .$ Let $\mathscr{U}$ be all the proper relatively open convex subsets of $K$. since $\mathscr{X}$ is a $\mathrm{LCS}$ and $K \neq \square$ (and let's assume that $K$ is not a singleton), $\mathscr{U} \neq \square$. Let $\mathscr{U}_{0}$ be a chain in $\mathscr{U}$ and put $U_{0}=\cup\left\{U: U \in \mathscr{U}_{0}\right\} .$ Clearly $U_{0}$ is open, and since $\mathscr{U}_{0}$ is a chain, $U_{0}$ is convex. If $U_{0}=K,$ then the compactness of $K$ implies that there is a $U$ in $\mathscr{U}_{0}$ with $U=K,$ a contradiction to the property of $U .$ Thus $U_{0} \in \mathscr{U} .$ By Zorn's Lemma, $\mathscr{U}$ has a maximal element $U$ If $x \in K$ and $0 \leqslant \lambda \leqslant 1$, let $T_{x, \lambda}: K \rightarrow K$ be defined by $T_{x, \lambda}(y)=\lambda y+(1-\lambda) x$ Note that $T_{x, \lambda}$ is continuous and $T_{x, \lambda}\left(\sum_{j=1}^{n} \alpha_{j} y_{j}\right)=\sum_{j=1}^{n} \alpha_{j} T_{x, i}\left(y_{j}\right)$ whenever $y_{1}, \ldots, y_{n} \in K, \alpha_{1}, \ldots, \alpha_{n} \geqslant 0,$ and $\sum_{j=1}^{n} \alpha_{j}=1 .$ (This means that $T_{x, \lambda}$ is an affine map of $K$ into $K .$ ) If $x \in U$ and $0 \leqslant \lambda<1,$ then $T_{x, \lambda}(U) \subseteq U .$ Thus $U \subseteq T_{x, \lambda}^{-1}(U)$ and $T_{x, \lambda}^{-1}(U)$ is an open convex subset of $K .$ If $y \in(\mathrm{cl} U) \backslash U, T_{x, \lambda}(y) \in[x, y) \subseteq U$ by Proposition IV.1.11. So cl $U \subseteq T_{x . \lambda}^{-1}(U)$ and hence the maximality of implies $T_{x, \hat{\lambda}}^{-1}(U)=K .$ That is $7.5 \quad T_{x . \lambda}(K) \subseteq U \quad$ if $\quad x \in U \quad$ and $\quad 0 \leqslant \lambda<1$
Claim. If $V$ is any open convex subset of $K,$ then either $V \cup U=U$ or $V \cup U=K$ In fact, (7.5) implies that $V \cup U$ is convex so that the claim follows from the maximality of $U$
It now follows from the claim that $K \backslash U$ is a singleton. In fact, if $a, b \in K \backslash U$ and $a \neq b,$ let $V_{a}, V_{b}$ be disjoint open convex subsets of $K$ such that $a \in V_{a}$
and $b \in V_{b} .$ By the claim $V_{a} \cup U=K$ since $a \notin U .$ But $b \notin V_{a} \cup U, a$ contradiction. Thus $K \backslash U=\{a\}$ and $a \in \operatorname{ext} K$ by $(7.3 e) .$ Hence $\operatorname{ext} K \neq \square$ Note that we have actually proved the following. 7.6 If $V$ is an open convex subset of $\mathscr{X}$ and ext $K \subseteq V,$ then $K \subseteq V$ Assume (7.6) is false. That is, assume there is an open convex subset $V$ of $\mathscr{X}$ such that ext $K \subseteq V$ but $V \cap K \neq K$. Then $V \cap K \in \mathscr{U}$ and is contained in a maximal element $U$ of $\mathscr{U}$. since $K \backslash U=\{a\}$ for some $a$ in ext $K$, this is a contradiction. Thus (7.6) holds. Let $E=\overline{\mathrm{co}}(\operatorname{ext} K) .$ If $x^{*} \in \mathscr{X}^{*}, \alpha \in \mathbb{R},$ and $E \subseteq\left\{x \in \mathscr{X}: \operatorname{Re}\left\langle x, x^{*}\right\rangle<\alpha\right\}=V$ then $K \subseteq V$ by $(7.6) .$ Thus the Hahn-Banach Theorem (IV.3.13) implies $E=K$.
why we define $ \quad T_{x . \lambda}$ ? why $T_{x, \lambda}$ is continuous ?
Unfortunately, I haven't fully understood the direction that this proof is going in, but continuity of T_{x,\lambda} is easy: X is a locally convex space, where addition and scalar multiplication are continuous maps. As T is a composition of these maps, it is also continuous.