This may remind you of a familiar correspondence theorem from group theory. If J is a subgroup of H, is a subgroup of G, G mod H is the same as G/J mod H/J. The theorems are related, if you dig down far enough.
Let a cover A consist of two sets, U and S-V. Naturally these cover S, but their interiors also cover S. This because the interior of S-V is S sans the closure of V, and the closure of V is contained in the interior of U. So the theorems regarding A-small chains apply.
Use the previous theorem to show h(S/U) is h(S/U) when restricted to A-small chains.
The A-small chains of U or S-V are precisely the chains of U or S-V. Any A-small complex in S has a piece in U and a piece in S-V. As a group, the A-small chains of S are generated by, though not necessarily a direct sum of, the chains of U and the chains of S-V.
When looking at the relative homology h(S/U), the chains of U contribute nothing. We are left with the chains of S-V, mod the chains of S-V intersect U. The kernel is now the chains of U-V, and that completes the proof.