Let $\{ U_k:k\in \mathbb{N}\}$ be an increasing sequence of open subsets of $\mathbb{R}^d$ whose union is $\mathbb{R}^d$ and such that each $K_k:=\overline{U_k}$ is compact and $K_k\subseteq U_{k+1}$.
We then have that $$ C_c^\infty (K_k)=\left\{ \phi \in C^\infty (\mathbb{R}^d):\mathrm{supp}[\phi ]\subseteq K_k\right\} $$ is a Fréchet space with the usual seminorms ($\sup$-norms of partial derivatives). We also have inclusion maps $i_k:C_c^\infty (K_k)\rightarrow C_c(\mathbb{R}^d)$.
On the other hand, we also have that $$ C^\infty (U_k)=\left\{ \phi :U_k\rightarrow \mathbb{C}:\phi \text{ is smooth.}\right\} $$ is a Fréchet space with the usual seminorms ($\sup$-norms of partial derivatives restricted to compact subsets). We also have restriction maps $r_k:C_c^\infty (\mathbb{R}^d)\rightarrow C^\infty (K_k)$.
Is $C_c^\infty (\mathbb{R}^d)$ to be equipped with the final topology with respect to the former sequence of maps (i.e. the direct limit of this sequence) or the initial topology with respect to the latter sequence of maps (i.e. the inverse limit of this sequence)? Perhaps the topology is the same?
EDIT: I believe I have answered this question (see my answer below), and this has provoked yet another question.
If I am correct, then, of the two choices presented above, we should equip $C_c^\infty (\mathbb{R}^d)$ with the direct limit topology of the sequence $C_c^\infty (K_k)$. This is indeed standard (at least, every source I have seen does it this way), but I want to know why. I first thought of the above alternative, and discounted it for reasons given in my answer. But what about this other 'obvious' alternative: for a fixed compact set $K\subseteq \mathbb{R}^d$ and non-negative integer $n$, define $$ p_{K,n}(f)=\sup \left\{ \left| \partial ^\alpha f(x)\right| :x\in K,|\alpha |\leq n\right\} , $$ and equip $C_c^\infty (\mathbb{R}^d)$ with the topology generated by this collection of seminorms. This should be the subspace topology when thought of as a subset of $C^\infty (\mathbb{R}^d)$, and so is quite natural. Why is it that this is not the topology conventionally equipped when $C_c^\infty (\mathbb{R}^d)$ is used as the space of test functions for distribution theory?
I believe I may have answered my question (now that it is well-posed). Please comment if you find an error in my logic.
At the end of the day, we may equip $C_c^\infty (\mathbb{R}^d)$ with any topology we please. In particular, we could equip it with the initial or final topologies mentioned in the question. However, of these two, only one of them makes $C_c^\infty (\mathbb{R}^d)$ into a limit.
We have an inclusion map $i:C_c^\infty (\mathbb{R}^d)\rightarrow C^\infty (\mathbb{R}^d)$ as well as other restriction maps $r_k':C^\infty (\mathbb{R}^d)\rightarrow C^\infty (U_k)$ such that $r_k=r_k'\circ i$. Hence, $C_c^\infty (\mathbb{R}^d)$ cannot be the inverse limit. My guess is that in fact $C^\infty (\mathbb{R}^d)$ is the inverse limit of this sequence, although I have not sat down and proved universality yet. Likewise, my guess is that $C_c^\infty (\mathbb{R}^d)$ is the direct limit of the former sequence (although, once again, I have not confirmed this).
Thus, while we can equip any topology we like, the 'correct' (in a categorical sense) is the direct limit topology of the sequence $C_c^\infty (K_k)$.
EDIT: With respect to the second part of my question, that is, why we do not equip $C_c^\infty (\mathbb{R}^d)$ with the subspace topology inherited from $C^\infty (\mathbb{R}^d)$, I asked my professor via e-mail the other day and he informed me that the reason is that because the resulting space is incomplete. A counterexample is more easily found in $C_c^\infty ((0,1))$ where you can construct a sequence of functions $f_n$ whose support is, say, $[1/n,1/2+1/n]$ that converges in $C^\infty ((0,1))$ but whose limit will not lie in $C_c^\infty ((0,1))$. For this reason, we instead equip $C_c^\infty (\mathbb{R}^d)$ with the direct limit topology, which is complete.