question. 1
Let $X$ be a locally compact Hausdorff space, and
$$
\mathcal{C}(X)=\{ f : X \rightarrow \mathbb{C} \mid f \ \mathrm{is \ continuous}\}
$$
$$
\mathcal{C}_c(X) = \{ f \in \mathcal{C}(X) \mid \mathrm{supp}f \mathrm{ \ is \ compact} \}
$$
and if $K \subset X$ is compact,
$$
\mathcal{C}(X,K) = \{ f \in \mathcal{C}(X) \mid \mathrm{supp}f \subset K \}
$$
Define the topology of $\mathcal{C}_c(X)$ as $\varinjlim_K \mathcal{C}(X,K)$.
($\varinjlim_K \mathcal{C}(X,K)$ means the colimt in $\mathbf{Top}$ of Banach spaces $\mathcal{C}_c(K)$)
Then, is $\varinjlim_K \mathcal{C}(X,K)$ topological vector space? How to prove?
question.2
Especially when $X= \mathbb{R}^n$, is $\varinjlim_K \mathcal{C}(X,K)$ topological vector space?
(I'm sorry my English is broken...I would be glad if you could help me.)
The answer to question 1 is "no".
Define $E(X) = \varinjlim_K \mathcal{C}(X,K)$ as the colimit in $\mathbf{Top}$. As a set we have $E(X) = \mathcal{C}_c(X)$, but the topology on $E(X)$ is finer than the norm topology on $\mathcal{C}_c(X)$.
Let $X$ be a discrete space which is clearly locally compact. Then the compact subsets of $X$ are precisely the finite subsets. It is readily seen that $E(X)$ carries the finite topology, i.e. $U \subset E(X)$ is open if and only $U \cap V$ is open in $V$ for each finite-dimensional linear subspace $V \subset E(X)$, where $V$ is given the standard Euclidean topology. This is true because each finite dimensional linear subspace is contained in some $\mathcal{C}(X,K)$, the latter being finite dimensional with the Euclidean topology.
It is well-known that that a vector space with the finite topology is never a TVS if its dimension is $\ge 2^{\aleph_0}$. Therefore, if $X$ is discrete with $\ge 2^{\aleph_0}$ elements, then $E(X)$ is no TVS.