While learning the concept of test functions, I am stuck on the following exercise. Following the book's notation we define $C^\infty_c (\mathbb{R}^d)$ as the set of smooth, compactly supported functions mapping from $\mathbb{R^d} \to \mathbb{C}$, and $C^\infty_c(K) \subset C^\infty_c(\mathbb{R}^d)$ the subset who are supported in $K$.
Let $K$ be a compact set in $\mathbb{R}^d$. Show that a linear map $T: C^\infty_c(K) \to X$ into a normed vector space $X$ is continuous if and only if there exists $k\geq 0$ and $C > 0$ such that $\|Tf\|_{K} \leq C \| f\|_{C^k}$ for all $f\in C^\infty_c(K)$.
Prior to this exercise my textbook defines the topology on $C_c^\infty (K)$ as the topology generated by the normS $$\| f\|_{C^k}=:\sup_{x\in K}\sum_{j=1}^k |\nabla^j f(x)|,$$
with $k = 1, 2, \dots$ and $\nabla^j f(x)$ being the $d^j$-dimensional vector.
I found much trouble proving the forward direction ($\implies$). Here is what I have tried.
Let $\tau_k$ be the topology generated by the norm $\|\cdot\|_{C^k}$. Since $\| f\|_{C^{k'}}\geq \|f\|_{C^k}$ whenever $k' \geq k$, we know that $\tau_k \subset \tau_{k'}$. It follows that the topology of $C_c^\infty(K)$ is given by $\cup_{i=1}^\infty \tau_k$. We will prove the forward direction if we can show that there exists $k>0$ such that for all open set $O$ in $X$, $T^{-1}(O)\in \tau_k$. As this will then imply $\|Tf\|_{K} \leq C \| f\|_{C^k}$ due to the equivalence between boundedness and continuity of an operator.
However, I cannot convince myself that such finite $k$ exists. Specifically, what if there is a sequence of $O_k$ open in $X$ such that $O_k\in \tau_k $ but $O_k \notin \tau_{k-1}$ for every $k>1$? If this possibility is true, then it would contradict the statement of the exercise.
I would appreciate it if someone could point out any conceptual error or provide hints to solving this exercise.
This is my answer to my own question after reading a reference suggested in the comment.
Assume continuity of $T$ with respect to the topology of $C^\infty_c(K)$. To show that the linear operator $T$ is continuous with respect to $\tau_k$ for some finite $k$, it suffices to show that there exists a fixed $k$ such that for all $r> 0$, there exists a $\delta>0$ satisfying $\{f: \|f\|_{C^k} < \delta\} \subset T^{-1}(B_r(0))$.
Since $\{\{f: \|f\|_{C^k}< \epsilon \} : k \geq 0, \epsilon > 0\}\subset \tau_k$ forms a neighborhood basis of $0$, there exists $k>0, \epsilon>0$ such that $N{\epsilon}(0) =: \{f: \|f\|_{C^k}< \epsilon\}\subset T^{-1}(B_1(0))$. Then for arbitrary $r> 0$, we get
$$ N_{r\epsilon}(0) := \{f: \|f\|_{C^k} < r\epsilon\}\subset T^{-1}(B_r(0)) $$
Hence $T$ is continuous at $0$ with respect to $\tau_k$.