Let $\Omega \subset \mathbb{R}^n$ be an arbitrary open set and $(x_n)_{n \in\mathbb{N}} \subset \Omega$ a sequence. Let $(a_n)_{n \in\mathbb{N}} \subset \mathbb{C}$ be a sequence such that $$\sum_{j=1}^{\infty} |a_j| < \infty.$$
I want to prove that the functional $T:C_c^0(\Omega) \longrightarrow \mathbb{C}$ given by $$T(\varphi)=\sum_{j=1}^{\infty} a_j\varphi(x_j),\; \forall \; \varphi \in C_c^0(\Omega)$$ is a Radon meausure in $\Omega$, that is, I want to prove that $T$ is continuous when $C_c^0(\Omega)$ is equipped with the topology inductive limit of the spaces $C_c^0(K)$, here $K \subset \Omega$ is an arbitrary compact subset of $\Omega$.
I thought of the following: it is enough I prove that $ T $ is continuous in $C_c^0(K)$, for all $K \subset \Omega$ compact, where the space $C_c^0(K)$ is (in particular) a Banach space with the norm $$|f|_K=\sup_{x \in K} |f(x)|, \; \forall \; f \in C_c^0(K).$$
That's what I thought true? This is the way?
Your thought is in the right direction.
You can also consider $b_n=\max(a_n,0)$ and $c_n=-\min(a_n,0)$.
Then $T_+\phi=\sum_n b_n\phi(x_n)$ and $T_- \phi=\sum_n c_n\phi(x_n)$. Both $T_+$ and $T_-$ are nonenagtive bounded operators on $C_{00}(\Omega)$. The Reisz-Markov representation implies that $T_+$ and $T_-$ can be represented as positive finite measures on $(\Omega,\mathscr{B}(\Omega))$. $T=T_+-T_-$ and $|T|=T_+ +T_-$ (this last part requires a small justification, but it is not complicated), thus $T$ is represented by a Radon measure ($|T|$ is the variation measure of $T$).