proof of Krein-Milman theorem

The proof is consist of three steps for good understanding.We will show initially that the set of extreme points of $K$,$Ex(K)$ is non-empty, $Ex(K)\ne \mathrm{\varnothing }$.We consider that $𝒜=\{A\subset K:A\subset K,$extreme$\}$.
Step1
The family set $𝒜$ ordered by $\subset$ has a minimal element, in other words there exist $A\in 𝒜$such as $\forall B\in 𝒜,B\subset A$ we have that $B=A$.
Proof1
We consider . The ordering relation is a partially relationon $𝒜$. We must show that $A$ is maximal element for $𝒜$.We apply Zorn’s lemma.We suppose that

$$𝒞=\{A_i:i\in I\}$$

is a chain of $𝒜$.Witout loss ofgenerality we take $A={\bigcap }_{i\in I}{A}_i$ and then $A\ne \mathrm{\varnothing }$. $𝒞$ has the property of finite intersectionsand it is consist of closed sets. So we have that ${\bigcap }_{i\in I}{A}_i\ne \mathrm{\varnothing }$. It is easy to see that $A\in 𝒜$.Also $A\subset A_i$, for any $i\in I$, so we have that $A>A_i$, for any $i\in I$.
Step2
Every minimal element of $𝒜$ is a set which has only one point.
Proof2
We suppose that there exist a minimal element $A$ of $𝒜$ which has at least two points,$x,y\in A$. There exist $x^{*}\in X^{*}$ such as $x^{*}(x)\ne x^{*}(y)$, witout loss ofgenerality we have that . $A$ is compact set (closed subset of the compact $K$). Also thereexist $\alpha \in ℝ$ such that $\alpha ={sup}_{z\in A}{x}^{*}(z)$ and $B=\{z\in A:x^{*}(z)=\alpha \}\ne \mathrm{\varnothing }$.It is obvious that $B$ is an extreme subset of $A$, $B$ is an extreme subset of $K$,$B\in 𝒜$.$x\notin B$ since $B\in 𝒜$ and $B⊊A$ that contradicts to the fact that $A$ is minimal extreme subset of $𝒜$.
From the above two steps we have that $Ex(K)\ne \mathrm{\varnothing }$.
Step3
$K=\overline{c}o(Ex(K))$ where $\overline{c}o(Ex(K))$ denotes the closed convex hull of extreme points of $K$.
Proof3
Let $L=\overline{c}o(Ex(K))$. Then $L$ is closed subset of $K$, therefore it is compact, and convex clearly by the definition.We suppose that $L⫋K$. Then there exist $x\in K-L$. Let use Hahn-Banach theorem(geometric form).There exist $x^{*}\in X^{*}$ such as . Let $\alpha =sup\{x^{*}(y):y\in K\}$, $B=\{y\in K:x^{*}(y)=\alpha \}$. Similar toStep2 $B$ is extreme subset of $K$. $B$ is compact and from step1 and step2 we have that $Ex(B)\ne \mathrm{\varnothing }$. It is true that$Ex(B)\subset Ex(K)\subset L$. Now let $y\in Ex(B)$ then $x^{*}(y)=\alpha$ and if . That is a contradiction.