Problem. Let $X$ be a topological spaces which is compact and Hausdorff, $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$, and suppose there exists a sequence $\{x_n\}_{n\in\mathbb N}\subset X$, such that $x_n\neq x_m$ for $n\neq m$. Moreover, assume that we have a sequence $\{\lambda_n\}_{n\in\mathbb N}\subset \mathbb{K}$ with the property that for every $f\in C(X,\mathbb{K})$ the series $\sum_{n\in\mathbb N}\lambda_n\, f(x_n)$ is convergent.
Show that $\sum_{n\in\mathbb N} \lvert\lambda_n\rvert<\infty$.
The opposite implication also holds. It follows from Weierstrass theorem.
We assume that for every $\,f\in C(X,\mathbb K)$, the series $\sum_{n\in\mathbb N}\lambda_n\,f(x_n)$ converges.
Define $$ T_n: C(X,\mathbb K)\to\mathbb K,\quad\text{as}\quad T_n(\,f)=\sum_{k=1}^n\lambda_k\,f(x_k). $$ Clearly, the $T_n$'s are linear functionals. Also, for every $\,f\in C(X,\mathbb K)$, the sequence $\,T_n(\,f)$ converges in $\mathbb K$, and hence $$ \sup_n \lvert T_n(\,f)\rvert<\infty. $$ Therefore, by virtue of the Uniform Boundedness Principle, $$ \sup_{n}\|T_n\|<\infty. $$ Now, it remains to show that $$ \|T_n\|=\sum_{k=1}^n\lvert\lambda_k\rvert. $$ The "$\le$" part of the above is straight-forward. Say now that $\lambda_n=w_n\lvert\lambda_n\rvert$, with $\lvert w_n\rvert=1$. We shall show that there exists an $f_n\in C(X,\mathbb K)$, with $f(x_i)=w_i$, for $i=1,\ldots,n$. As $X$ is $T_2$, then there exist $U_1,\ldots,U_n$, open and disjoint neighbourhoods of $x_1,\ldots,x_n$, respectively. As $X$ is compact and Hausdorff, it is also completely regular, and thus, it is possible, for $i=1,\ldots,n$, to define $$ g_i :X\to\mathbb R, $$ such that $g_i(x_i)=1$, $0\le g_i(x)\le 1$, for all $x\in X$ and $g_i(x)=0$, for $x\in X\setminus U_i$.
Then the sought for $f_n$ can be defined as $f_n=\sum_{i=1}^n w_ig_i$. Clearly, $\|f_n\|=1$, and $T_n(f_n)=\sum_{i=1}^n\lvert f_i(x_i)\rvert=\sum_{i=1}^n\lvert \lambda_i\rvert$. Thus we finished with the "$\le$" part as well.