The space of all infinitely differentiable functions with compact support in region $\Omega \subset \mathbb{R}^n$, is not trivial for any $\Omega$, and it is a vector space over $\mathbb{R}$, I denote it with $D(\Omega)$(I do not here care about topology)(who likes to distinguish things let him/her take $C_0^{+\infty}$ or $C_c^{\infty}(\Omega)$, that is not the point here), and, like any other vector space, it must have a basis and a dimension. So, the most normal question would be what is the dimension of that vector space.
My guess, which I do not know if it is true, is that it should be infinity, because if we take $(a,b)$(for example) and now for some $(a_1,b_1)\subset (a,b)$, $(a_1,b_1) \neq (a,b)$, there is an $\epsilon > 0$ , where $\epsilon = max\{|a-a_1|,|b-b_1|\}$, and now we can take function $f$ to be any polynomial in $[a_1,b_1]$, that function $f$ satisfies that for $x \in (a,a_1)$ holds $f(x)<f(a_1)$, and for $x \in (b_1,b)$, it holds $f(x) < f(b_1)$, $f$ is on $(a,a+\frac{\epsilon}{3})$ and on $(b-\frac{\epsilon}{3}, b)$ always zero, and $f \in C^{\infty}(a,b)$, and this function exists and is in $D(a,b)$, so basically this is why it should be infinity. But, I do not know if any of this has any nonempty intersection with something which would be true. Thanks for help!
There is a standard result that for any real sequence $\{a_n\}$ there exists a function $\varphi\in D(\Bbb R)$ such that $$\forall n\ge 0 \quad \varphi^{(n)}(0)=a_n.$$ By a slight modification of this result you can see that the space of test functions on any non-empty open interval is infinite-dimensioned.
A somewhat harder exercise would be to show that the basis of this space has cardinality of at least continuum (the argument above shows that the basis is at least countable).