$C^\infty_c (K)$ is separable for $K$ compact $C^\infty_c (K)$ is the space of smooth functions supported on $K$ a compact subset of $\mathbb{R}^d$. For simplicity assume $K$ is just a ball centered at the origin. This has the smooth topology in which convergence is uniform convergence of the functon and all derivatives. 
Note that convergence in the smooth topology of $C^\infty_c(K)$ is the same as uniform convergence of all derivatives. 
I'm trying to show this space is separable. I initially wanted to use polynomials with rational coefficients, as these are countable and dense in $C_c(K)$ with uniform topology. Then using the fundamental theorem of calculus, we can easily show that this set is also dense in the smooth topology.
But the polynomials are not supported on $K$ so this fails. I was thinking about multiplying by a smooth function supported on $K$ that equals $1$ on most of $K$, but then I'm not sure if the fundamental theorem of calculus argument will still work. 
So what can I do?

Instead of $K$ compact, take $K$ bounded and open (clearly $C^\infty_c(K)$ is dense in $C^\infty_c(\overline{K})$)
As you said the set of polynomials is separable and is dense in $C^\infty(\overline{K})$, but the polynomials are not $C^\infty_c(K)$. 
Consider a sequence of open sets $\Omega_n \subset \Omega_{n+1} \subset \ldots \subset K$ such that any $\varphi \in C^\infty_c(K)$  has its support contained in one of those $\Omega_n$.
Take $\phi_n \in C^\infty_c(K)$ such that $\phi_n = 1$ on $\Omega_n$. 
For any $\varphi \in C^\infty_c(K), \text{supp}(\varphi) \subset \Omega_n$, we have  a polynomial sequence $p_k$ such that $\lim_k p_k = \varphi$ in  $C^\infty(K)$, thus $\lim_k p_k \phi_n = \varphi \phi_n= \varphi$ in  $C^\infty_c(K)$.
Qed $C^\infty_c(K)$ is the closure of those $p \phi_n$ and is separable.
