Let $\Omega\subseteq\mathbb{R}^n$ be some nonempty open, and use the notation $U\Subset V$ to imply that $U$ is a compact subset of $V$. Then, for all $K\Subset\Omega$, we can define the space $$\mathscr{C}^\infty_K(\Omega)=\{\varphi\in \mathscr{C}^\infty(\Omega)\mid \operatorname{supp}\varphi\subseteq K\}$$ with the seminorms $\lVert f\rVert_{\alpha,K}=\lVert\partial^\alpha f\rVert_{L^\infty(\Omega)}$, so that $$\{\mathscr{C}^\infty_K(\Omega)\}_{K\Subset\Omega}$$ is an inductive system of topological spaces with maps given by inclusion as subspaces, so we can define the inductive limit of this system of $\mathscr{C}^\infty_0(\Omega)$ with the inductive limit topology, so that for any inclusion $K\Subset K'\Subset \Omega$ the inclusion $$\mathscr{C}^\infty_K(\Omega)\subset\mathscr{C}^\infty_0(\Omega)$$ is simply the chain of inclusions $$\mathscr{C}^\infty_K(\Omega)\subset\mathscr{C}^\infty_{K'}(\Omega)\subset\mathscr{C}^\infty_0(\Omega).$$

Now, I wish to show that this is equivalent to the usual definition of $\mathscr{C}^\infty_0(\Omega)$ using convergence, i.e. that for $\varphi_n,\varphi\in\mathscr{C}^\infty_0(\Omega)$, we have that $$\varphi_n\xrightarrow{n\to\infty}\varphi$$ in $\mathscr{C}^\infty_0(\Omega)$ if and only if $\varphi_n,\varphi$ all have support in some $K\Subset\Omega$, and $$\varphi_n\xrightarrow{n\to\infty}\varphi$$ in $\mathscr{C}^\infty_K(\Omega)$. Now, since $\mathscr{C}^\infty_K(\Omega)$ is a subspace of $\mathscr{C}^\infty_0(\Omega)$ it is trivial to see that convergence in the smaller space implies convergence in the larger one. However, I don't see how we can conclude that if $$\varphi_n\xrightarrow{n\to\infty}\varphi$$ in $\mathscr{C}^\infty_0(\Omega)$, we can restrict to some $K\Subset\Omega$. I see that $\mathscr{C}^\infty_K(\Omega)\subset\mathscr{C}^\infty_0(\Omega)$ is a closed subspace for all $K\Subset \Omega$, but I can't find any neighborhood of $\varphi$ that would help me restrict to some $\mathscr{C}^\infty_K(\Omega)$.