Fix $\Omega$ as a non-empty open subset of $\mathbb{R}^{d}$ throughout. Also let $\mu$ denote the Lebesgue measure on $\mathbb{R}^{d}$.
You are correct in pointing out that $C_{c}(\Omega )$ is dense in $L^{p}(\Omega )$ while being an uncountable set. One way to see that $C_{c}(\Omega )$ is uncountable is that if $x_{0} \in \Omega$ is fixed, then for each $\alpha \in \mathbb{R}$ there exists some $f\in C_{c}(\Omega )$ such that $f(x_{0}) = \alpha$. This obtains an injection of the uncountable set $\mathbb{R}$ into $C_{c}(\Omega )$.
However, as mentioned in the comments, it is not necessary for $C_{c}(\Omega )$ to be a countable set. The main idea is to look at a smaller dense space where it is easier to find a countable dense subset. Then such a countable subset will also be dense in the larger space.
We will modify this proof to obtain the following result.
Theorem. Let $p\in [1, \infty )$. Then $C_{c}(\Omega )$ is separable when equipped with the $L^{p}$-norm.
Proof. Let $(K_{n})_{n\in\mathbb{N}}$ be a sequence of compact subsets of $\Omega$ such that $K_{n} \subseteq {\rm int} \, K_{n+1}$ for all $n\in\mathbb{N}$ and $\bigcup_{n\in\mathbb{N}} K_{n} = \Omega$. To see that such a sequence exists, you may take
\begin{equation}
K_{n} = \{x\in \Omega : \|x\| \leq n \; \text{and} \; {\rm dist}(x, \mathbb{R}^{d}\setminus \Omega ) \geq 1/n \}
\end{equation}
for each $n\in\mathbb{N}$ if desired. Such a sequence $(K_{n})_{n\in\mathbb{N}}$ has the additional property that if $K\subseteq \Omega$ is compact, then $K \subseteq K_{n}$ for some $n\in\mathbb{N}$. To see this, note that $\{ {\rm int} \, K_{m} : m \in \mathbb{N} \}$ covers $\Omega$. In particular, this collection of open sets covers the compact set $K$, so a finite collection of them cover $K$. Since the collection is increasing in $m$, there is some $n\in\mathbb{N}$ such that $K \subseteq {\rm int} \, K_{n} \subseteq K_{n}$ as desired.
Next, for each $n\in\mathbb{N}$ define $C_{n} := \{ f\in C_{c}(\Omega ) : {\rm supp} \, f \subseteq K_{n} \}$. We claim $C_{c}(\Omega ) = \bigcup_{n\in\mathbb{N}} C_{n}$. Clearly $\bigcup_{n\in\mathbb{N}} C_{n} \subseteq C_{c}(\Omega )$. For the other inclusion, let $f\in C_{c}(\Omega )$. Then ${\rm supp} \, f$ is compact and so is contained in $K_{n}$ for some $n\in\mathbb{N}$. Hence $f\in C_{n} \subseteq \bigcup_{n\in\mathbb{N}} C_{n}$.
Fix $n\in\mathbb{N}$. Define $\phi_{n} : C_{n} \to C(K_{n})$ by $\phi_{n}(f) := f\vert_{K_{n}}$. Since ${\rm supp} \, f \subseteq K_{n}$ for each $f\in C_{n}$ we have that $\phi_{n}$ is an isometry from $(C_{n}, \|\cdot\|_{\infty} )$ into $(C(K_{n}), \|\cdot\|_{\infty} )$. Because $K_{n}$ is a compact metrisable space we have that $C(K_{n})$ is separable. Hence $C_{n}$ is isometrically isomorphic to a subspace of a separable space, so $C_{n}$ is also separable with the uniform norm. Consequently, for each $n\in\mathbb{N}$ there exists a countable set $A_{n} \subseteq C_{n}$ which is dense in $(C_{n}, \|\cdot\|_{\infty} )$.
Now define $A := \bigcup_{n\in\mathbb{N}} A_{n}$. Then $A$ is countable. We claim $A$ is dense in $(C_{c}(\Omega ), \|\cdot\|_{L^{p}})$. Let $f\in C_{c}(\Omega )$ and $\varepsilon > 0$. Then there is some $n\in\mathbb{N}$ such that ${\rm supp} \, f \subseteq K_{n}$. Choose $g\in A_{n}$ such that $\|g - f\|_{\infty} < \varepsilon / (1 + \mu (K_{n})^{1/p})$. (At this point it is being used that the Lebesgue measure is finite on compact sets.) Then $g\in A$ and it follows, since ${\rm supp}(g-f) \subseteq K_{n}$, that $\|g - f\|_{L^{p}} \leq \mu (K_{n})^{1/p} \|g - f\|_{\infty} < \varepsilon$. Therefore $A$ is a countable dense subset of $(C_{c}(\Omega ), \|\cdot\|_{L^{p}} )$ and this completes the proof.
