Norbert pointed out in a comment that this is false for real scalars, so we assume that we consider complex scalars $C(X) = C(X,\mathbb{C})$.
I haven't thought about it deeply, so maybe there are easier approaches. We are going to apply the Stone-Weierstrass theorem.
From the description of extreme points the following are straightforward:
- If $f$ is extremal then so are its complex conjugate $\bar{f}$ and $e^{i\alpha} f$ for $\alpha \in \mathbb{R}$.
- If $f$ and $g$ are extremal then $f\cdot g$ is extremal.
- If $f$ and $g$ are convex combinations of extreme points then so is $\frac{1}{2}(f+g)$.
If $f$ and $g$ are convex combinations of extreme points then so is $f \cdot g$.
To see the last point, let $f = \sum_{j=1}^m \lambda_j f_j$ and $g = \sum_{k=1}^n \mu_k g_k$ with $\sum \lambda_j = 1 = \sum \mu_k$ and $\lambda_j, \mu_k \geq 0$ and $f_j, g_k$ extremal. Then $f \cdot g = \sum_{j=1}^m \sum_{k=1}^n (\lambda_j \mu_k) \, f_j g_k$ with $\sum_{j=1}^m \sum_{k=1}^n \lambda_j \mu_k = \big( \sum_{j=1}^m \lambda_j \big) \big(\sum_{k=1}^n \mu_k\big)= 1$ and $\mu_j \lambda_k \geq 0$ and $f_j g_k$ is extremal by 2.
It follows from this that the linear span of the extreme points (= the positive multiples of convex combinations of the extreme points by the above) is a self-adjoint subalgebra $A$ of $C(X)$ and since $A$ contains the constant functions, it remains to prove that elements of $A$ separate points of $X$:
If $X$ is empty or has only one point, the statement in the question is very easy to prove. So, let $x_1,x_2 \in K$ be two distinct points. Choose a continuous function $f \colon X \to [0,1]$ such that $f(x_1) = 0$ and $f(x_2) = 1$. Set $g(x) = 1 + i f(x)$. Then $g$ vanishes nowhere, so $h = \frac{g}{\lvert g\rvert}$ is well-defined, continuous and an extreme point of the unit ball. We have $h(x_1) \neq h(x_2)$, so $A$ separates points.
We can now apply Stone-Weierstrass theorem to $A$, showing that $A$ is dense in $C(X)$. From this it follows easily that $A \cap C(X)_1$ is dense in $C(X)_1$.