Is the space of compactly supported $k$-times differentiable functions separable? Is the space $C^k_c(\mathbb{R}^n)$ of compactly supported $k$-times continuously differentiable functions with the norm $\|f\|:=\sum_{|\alpha|\leq k}\|D^\alpha f\|_\infty$ separable?
 A: Yes, $C_c^k(\mathbb{R}^n)$ is separable with that norm.
By the Stone-Weierstraß theorem, the polynomials span a dense subspace of $C(S^n)$, where $S^n \subset \mathbb{R}^{n+1}$ is the $n$-dimensional unit sphere. The countable subset of polynomials with rational coefficients is also dense, so $C(S^n)$ is separable in the topology of uniform convergence. Among metric spaces, separability is inherited by subspaces (for metric spaces, separability is equivalent to second countability).
Thus $V = \{ f \in C(S^n) : f(0,\dotsc,0,1) = 0\}$ is a separable subspace. But under stereographic projection, $V$ is isometrically isomorphic to $C_0(\mathbb{R}^n)$, the space of continuous functions on $\mathbb{R}^n$ vanishing at infinity.
Second countability is preserved by finite products, so $C_0(\mathbb{R}^n)^{N(k)}$ is separable, where $N(k)$ is the number of multiindices with $\lvert\alpha\rvert\leqslant k$. The two norms
$$\lVert (f)\rVert_1 = \sum_{j=1}^{N(k)} \lVert f_j\rVert_\infty \text{ and } \lVert (f)\rVert_2 = \max \left\{ \lVert f_j\rVert_\infty : 1 \leqslant j \leqslant N(k)\right\}$$
are easily seen to be equivalent, and the latter obviously defines the product topology.
The space $C_0^k(\mathbb{R}^n)$ of $k$ times continuously differentiable functions on $\mathbb{R}^n$ such that all derivatives of order $\leqslant k$ vanish at infinity embeds isometrically as a closed subspace into $C_0(\mathbb{R}^n)^{N(k)}$ when it is endowed with
$$\lVert f\rVert = \sum_{\lvert\alpha\rvert \leqslant k} \lVert D^\alpha f\rVert_\infty,$$
and the product with $\lVert\,\cdot\,\rVert_1$. Hence it is separable.
Since separability is inherited by subspaces of metric spaces, the subspace $C_c^k(\mathbb{R}^n) \subset C_0^k(\mathbb{R}^n)$ is also separable under the given norm.
Note: $C_c^k(\mathbb{R}^n)$ is a dense proper subspace of $C_0^k(\mathbb{R}^n)$ (which is its completion), so it is not a Banach space.
