While reading this post, I was a bit confused by the following :
If $X$ is a compact metric space, then by Urysohn's Lemma and Stone-Weierstrass, the continuous functions $C(X)$ on $X$ are separable and hence the result follows as $C_{c}(X) = C(X)$.
I understand the use of Stone-Weierstrass because polynomial with rational coefficient are countable and are dense in polynomials with real coefficients which are dense (by stone-Weierstrass) in $C(X)$ when $X$ is compact. But I don't understand where Urysohn's Lemma comes into play.
Since $X$ is compact metric, it has a countable basis $\{U_n\}_{n\in {\mathbb N}}$ of open sets.
For each pair $(n, m)$ such that $\overline U_n\subseteq U_m$, you can use Urysohn to pick some $f_{n, m}$ in $C(X)$ such that $f_{n, m}=1$ on $U_n$, and $f_{n, m}=0$ on $X\setminus U_n$.
The subalgebra $A\subseteq C(X)$ generated by all of the $f_{n, m}$ separates points of $X$, so it is dense by Stone-Weierstrass. Now you can finish your argument by pretending that $A$ is the algebra of polynomials.