Note that $a_n \in \mathbb{R}$ for all n and $a\in \mathbb{R}$ as well.
Here's my attempt:
Let $S$ be a compact subset of $(a-R,a+R)$. Let $r:= \max \{|a-\inf S|, |a-\sup S|\} $. Since $\inf S \in S$, $|\inf S - a| <R$ and likewise $|\sup S - a|<R$. It is also clear that $\sum a_n r^n$ converges absolutely.
Now, let $x\in S$. It is easy to show that $|x-a|\le r.$ Now, for $n>m$, we have that $|a_m(x-a)^m + \ldots+a_n(x-a)^n|\le |a_mr^m|+\ldots+|a_nr^n|$. The right hand side of the inequality can be made arbitrarily small because $\sum a_n r^n$ converges absolutely. This completes the proof.
Is this proof correct? Alternative proofs are welcome.