If you are willing to quote the Schoenflies theorem and the classification of surfaces, a quick proof of this result is a standard exercise. First, using algebraic topology (as in the proof of the Jordan curve theorem itself), one shows that $S^1 - (A \cup B)$ is a union of three components, whose closures are three compact subsets $S_A$, $S_{AB}$, $S_B$ with frontiers $A$, $A \cup B$, $B$ respectively. Applying the Schoenflies theorem, each of $S_A$, $S_{AB}$, $S_B$ is a compact surface with boundary, and in fact $S_A$, $S_B$ are homeomorphic to a disc. Next one has $\chi(S_A) + \chi(S_{AB}) + \chi(S_B) = \chi(S^2)$, which is a standard result about gluing together a bunch of compact surfaces-with-boundary by pairwise identifying boundary circles. So that equation becomes $1 + \chi(S_{AB}) + 1 = 2 \implies \chi(S_{AB})=0$. Using the classification of surfaces, and ruling out all compact surfaces of Euler characteristic zero with other than two boundary components, $S_{AB}$ is an annulus.

The same technique carried out in general allows one to take any compact surface $S$ with boundary (specified by its Euler characteristic, orientability, and number of boundary components), any integer $n$, and any subset $\mathcal{C} \subset S$ consisting of $n$ disjointly embedded circles, and enumerate all of the finitely many possibilities for the topological types of the components of $S-\mathcal{C}$.