Yes, and the condition $0<\inf_n\|f_n\|$ isn't needed. Let $f$ denote the limit. Then for each $N\in \mathbb N$, $\left\|f-\sum\limits_{n=1}^N a_nf_n\right\|=\left\|\sum\limits_{n={N+1}}^\infty a_nf_n\right\|=\sup\limits_{n\geq N+1} |a_n|\|f_n\|,$
and this converges to $0$ as $N\to\infty$ because $(f_n)_n$ is uniformly bounded and $(a_n)$ converges to $0$. (Note how disjointness of the supports is used to infer that these series are well-defined and that the last equality is true.)
Answer to the original version, before the condition of uniform boundedness was added (I had posted after the edit but before I saw the edit):
Not in general. The series need not even represent a continuous function, let alone converge uniformly. E.g. let $K=\{0\}\cup\{\frac{1}{n}:n\in\mathbb Z^{>0}\}$ as a subspace of $\mathbb R$ with the usual topology, let $f_n=n^2\chi_{\{1/n\}}$, and let $a_n=\frac{1}{n}$.