$\mathcal{E}(C(K, \mathbb{C})) = \{\omega_a : a \in K\}$ We know that $C(K, \mathbb{C})$ is a $C^*$-algebra of complex valued functions on a compact $K$. Now, for all $a \in K$, define the linear functional $\omega_a \colon C(K, \mathbb{C}) \rightarrow \mathbb{C}$ by $\omega_a(f) = f(a)$. Then, $\omega_a$ is a state (it's not difficult to check that). 
Denote by $\Sigma(C(K, \mathbb{C}))$ the set of states of $C(K, \mathbb{C})$ and by $\mathcal{E}(C(K, \mathbb{C}))$ the set of pure states (ie, extremal states) of $C(K, \mathbb{C})$.
Now, define the following set subset of $C(K, \mathbb{C})$:
$C_a(K,\mathbb{C}) = \{ f \colon K \rightarrow \mathbb{C} : f(a) = 0 \}$.
Using the fact that $C(K, \mathbb{C}) = C_a(K, \mathbb{C}) \bigoplus \mathbb{C}\textbf{1}$, I need to show that:
$$\mathcal{E}(C(K, \mathbb{C})) = \{\omega_a : a \in K\}.$$
Part 1.   $\{\omega_a : a \in K\} \subset \mathcal{E}(C(K, \mathbb{C})).$
Suppose that $w_a$ is not a pure state. That means that we can write $w_a = \lambda \omega_1 + (1 - \lambda)\omega_2$, with $\omega_1, \omega_2 \in \Sigma(C(K, \mathbb{C}))$ and $\lambda \in (0, 1)$. 
Let $f \in C(K, \mathbb{C})$. So, $f = f_a + c,$ where $f_a \in C_a(K, \mathbb{C})$ and $c \in \mathbb{C}$. Then, 
$$\omega_a(f) = \omega_a(f_a) + c = 0 + c = c.$$
On the other hand,
\begin{align*} \omega_a(f) &= \lambda \omega_1(f) + (1 - \lambda)\omega_2(f)\\
&= \lambda(\omega_1(f_a) + c) + (1 - \lambda)(\omega_2(f_a) + c)\\
&= \lambda\omega_1(f_a) + (1 - \lambda)\omega_2(f_a) + c.
\end{align*}
Putting the above expressions together, we get that $\lambda\omega_1(f_a) = -(1-\lambda)\omega_2(f_a)$, which leads us to an absurd, because $\lambda > 0, (1 - \lambda) > 0, \omega_1(f_a) \geq 0$ and $\omega_2(f_a) \geq 0$.
Part 2. $\mathcal{E}(C(K, \mathbb{C})) \subset \{\omega_a : a \in K\}$.
This is where I'm struggling. Could someone help me? 
 A: I'm not aware of a totally elementary argument. Let  $S(K)$ denote the state space of $C(K)$. Here is a  fairly basic argument:


*

*Prove that if $X$ is a topological vector space and $A,B\subset M$ are both convex and compact, then $$ \overline{\operatorname{conv}(A\cup B)}=\operatorname{conv}(A\cup B).$$
We only need to show that inclusion $\subset$. Define $Y=[0,1]\times A\times B$ and $f:Y\times A\times B\to\operatorname{conv}(A\cup B)$ by $f(t,a,b)=ta+(1-t)b$. As $f$ is continuous and its domain is compact, the image of $f$ is compact; and it is also convex (exercise). Then 
$$
\overline{\operatorname{conv}A\cup B}\subset f(Y)\subset \operatorname{conv}(A\cup B).
$$

*Use the above to prove that if $X$ is a locally convex vector space, $Y\subset X$ is compact, and $\overline{\operatorname{conv}Y}$ is compact, then the extreme points of $\overline{\operatorname{conv} Y}$ are in $Y$. Indeed, suppose that $p\in\overline{\operatorname{conv}Y}\setminus Y$. As $Y$ is closed, there exists $V'$, open, with $0\in V'$ and $(p+V')\cap Y=\emptyset$. As $X$ is locally convex, we may find $V\subset V'$, open, convex, with $0\in V$, and such that $V-V\subset V'$. From this we obtain that $(p+V)\cap (Y+V)=\emptyset$. Then $p\not\in\overline{Y+V}$. Since $Y\subset _{y\in Y}(y+V)$, there exist by compactness $y_1,\ldots,y_n\in Y$ with $Y\subset \bigcup_{j=1}^n (y_j+V)$. Let $Y_j=\overline{\operatorname{conv}\overline{(y_j+V)}\cap Y}$; these sets are convex and compact. By the above
$$
\overline{\operatorname{conv} Y}=\overline{\operatorname{conv} Y_1\cup\cdots\cup Y_n}=\operatorname{conv} Y_1\cup\cdots\cup Y_n.
$$
Then $p-\sum_{j=1}^n\alpha_j y_j$ with $y_j\in Y_j$, $\alpha_j\geq0$, $\sum\alpha_j=1$. If $p$ were extreme, $p=y_k$ for some $k$. Then
$$
p\in Y_k\subset \bigcup_{j=1}^n Y_j\subset \bigcup_{j=1}^n\overline{y_j+V}\subset \overline{Y+V},
$$
a contradiction. So $p$ is not extreme.

*Prove that any state on $C(K)$ is a weak$^*$-limit of convex combinations point-evaluations. This one is very easy: if $\varphi$ is a state on $C(K)$ that is not in $\overline{\operatorname{conv}\,\{\omega_a:\ a\in K\}}^{w^*}$, then we can use Hahn-Banach (applied to the vector space $C(K)^*$ with the weak$^*$-topology, which is reflexive) to get $c\in \mathbb R$ and $f\in C(K)$ such that 
$$
\operatorname{Re}\omega_a(f)<c<\operatorname{Re}\varphi(f)
$$
for all $a\in K$. So we have 
$$
\operatorname{Re}f(a)+\|f\|<c+\|f\|<\operatorname{Re}\varphi(f)+\|f\|.
$$
We may rewrite this as 
$$
\operatorname{Re}(f+\|f\|)(a)<c+\|f\|<\operatorname{Re}\varphi(f+\|f\|)=\varphi(\operatorname{Re} f+\|f\|).
$$
As this works for all $a$ we get 
$$
\|\operatorname{Re}( f+\|f\|)\|<c+\|f\|<\|\operatorname{Re}(f+\|f\|)\|,
$$
a contradiction. 

*Now we collect the information:  we proved above that $S(K)=\overline{\operatorname{conv}\{\omega_a:\ a\in K\}}^{w^*}$. By the second bullet point, all extreme points of $S(K)$ are in $\{\omega_a:\ a\in K\}$. 
