$C^\infty_c (K)$ 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?
 A: I'll outline one possible answer here. Sorry, no time to make it very detailed.
Start with a bump function: A nonnegative $C^\infty$ function $\rho$ with support inside the unit ball centered at the origin, and integral $1$. Let $\rho_\delta(x)=\delta^{-d}\rho(x/\delta)$. Now the idea is to note that, if $f\in C_c^\infty(K)$ has its support in the interior of $K$ (which is good enough – use a cutoff function), then the convolution $f*\rho_\delta$ is in $C_c^\infty(K)$ when $\delta$ is small enough, and it converges to $f$ as $\delta\to0$. In fact the convergence is uniform, for
$$ f(x)-f*\rho_\delta(x)=\int_{\mathbb{R}^d} \bigl(f(x)-f(x-y)\bigr)\rho_\delta(y)\,dy, $$
so taking absolute values, using the triangle inequality, and noting the uniform continuity of $f$ ensures this is uniformly small when $\delta$ is small.
The identity $D^\alpha(f*\rho_\delta)=(D^\alpha f)*\rho_\delta$ lets us apply the above to all higher derivatives of $f$ as well.
Now limit your attention to rational $\delta$, and replace the integral of the convolution with Riemann sums. Conclude (once more using the uniform continuity of $f$) that $f$ (and its higher derivatives, again using the above trick) can be approximated arbitrarily well by linear combinations (with rational coefficients) of functions $\rho_\delta({\cdot}-a)$, with rational $\delta>0$ and $a\in\mathbb{Q}^d$. Since there are a countable number of these functions, we are essentially done.
(Edited to supply some more detail.)
A: For simplicity let's assume $K=$ the closed Unit ball in $R^n.$ Let $\lambda_n$ be a sequence of  positive number such that $\lambda_n ‎\nearrow‎ 1.$ 
Define 
$$A_{n} = \{P : K \overset{smooth}{\longrightarrow} R ~|  P ~\text{is a rational polynomials on}~ B_ {\lambda_n} \text{and } P=0 ~\text{on}~ K\setminus B_{\lambda_{n+1}}   \}$$
This set is well-defined by picking suitable bump-function (cut-off), we also may assume $A_n$ is countable by selecting only one $P$ satisfying RHS conditions in the set. Clearly $A_{n} \subset C^\infty_c (K).$ we show that $\bigcup_{n \in N} A_n$ is dense in $C^\infty_c (K)$. To end this pick $f \in C^\infty_c (K)$ and $\epsilon > 0.$ One can find enough large $n \in N$ such that first $\sup_{x \in K\setminus B_{\lambda_{n+1}}} |f (x)| \leq \frac{\epsilon}{4} $ and second $\sup_{x \in B_{\lambda_{n+1}}\setminus B_{\lambda_{n}} } |f (x)| \leq \frac{\epsilon}{4}$  (These two are possible since $f$ is close to zero near the boundary of $K$ is  ) Now choose $P \in A_{n}$ such that First $\sup_{x \in  B_{\lambda_{n}}}|f(x) - P (x)| \leq \frac{\epsilon}{4}$ (wirestrass theorem) and Second $\sup_{x \in B_{\lambda_{n+1}}\setminus B_{\lambda_{n}} } |P (x)| \leq \frac{\epsilon}{4}$ ($P$ near boundary of $K$ is close to zero) .  Then
$$ \| f- p \| \leq \sup_{x \in  B_{\lambda_{n}}}|f(x) - P (x)| + \sup_{x \in B_{\lambda_{n+1}}\setminus B_{\lambda_{n}} } |f (x)|+ \sup_{x \in B_{\lambda_{n+1}}\setminus B_{\lambda_{n}} } |P (x)|+\sup_{x \in K\setminus B_{\lambda_{n+1}}} |f (x)-0| \leq \epsilon $$    
A: 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.
