What set $Y$ that $C_{c}(U) \subset Y$ and $Y$ separable w.r.t. $\vert \vert \cdot \vert \vert_{\infty}$ Let $U \in \mathbb R^{d}$ open and further $C_{c}(U)$ be the space of functions with compact support. Show that $C_{c}(U)$ is separable w.r.t. $\vert\vert \cdot \vert \vert_{\infty}$, and I have been given the tip that:
$X$ is subset of separable metric space $(M,d)$, then $X$ is also separable. This is of course logical, but which set $M$ can I find that is separable. I've looked at the candidates $C(U), C_{b}(U)$ and $L^{\infty}(U)$ and none of them hold for separability w.r.t. the essential supremum. 
My main goal is to show that $C_{c}(U)$ is separable w.r.t. $\vert \vert \cdot \vert \vert_{p}$ for all $p \in [1,\infty[$ and I have been advised to go about it the way above, but I am unsure how proving separability w.r.t. $\vert \vert \cdot \vert \vert_{\infty}$ can help me in anyway. 
Any ideas, and hints are greatly appreciated. 
 A: The person who gave you the hint is probably thinking of $M$ being the space $C_0(U)$ consisting of continuous functions that vanish at $\partial U$ and at infinity (if $U$ is unbounded).  This is a Banach space and is well known to be separable.  However, proving this isn't really any easier than proving the separability of $C_c(U)$ directly.
Here's a sketch as to how you might prove this directly:


*

*Construct an "exhaustion" sequence of open sets $U_n \subset U$ such that $\overline{U_n}$ is compact and contained in $U_{n+1}$, and $\bigcup_n U_n = U$.  

*Use Urysohn's lemma to construct a sequence of functions $\phi_n$ such that $\phi_n = 1$ on $\overline{U_n}$ and is compactly supported in $U_{n+1}$.

*Consider the set of all functions of the form $\phi_n f$, where $f$ is a polynomial in $x_1, \dots, x_d$ with rational coefficients.  Show that this set is countable.

*Given $g \in C_c(U)$, find a sequence of functions from the above set which converge uniformly to $g$.  (Use the Weierstrass approximation theorem.)
A: *

*First, show that $C(X)$ with the norm $\| \cdot \| _\infty$ is separable whenever $X$ is compact. Almost full proofs are to be found here and here.

*Next, use (1.) and a compact exhaustion of $U$ to show that $C_c (U)$ with the norm $\| \cdot \| _\infty$ is separable, as shown here.

*Next, use the well-known result that $C_c (U)$ is dense in $L^p (U)$ for $p \in (1, \infty)$. Since $C_c (U)$ is separable, and $L^p (U)$ is first-countable (because is is a metric space), it follows that $L^p (U)$ is separable (the dense countable subset being the same one that is dense in $C_c (U)$).

*Finally, $C_c (U)$ with the norm $\| \cdot \| _p$ may be seen as a subspace of $L^p (U)$. Since the latter has been seen to be separable (and a metric space), so will be the subspace $(C_c (U), \| \cdot \|_p)$.
