Does $A\subset(\overline A)^0$ hold in this situation?
It's simple, really:
If $U_j$ is any of those intersecting $K_n$ (in the union at hand), by definition: $$\overline{U_j} \subseteq K_{n+1}$$ because $K_{n+1}$ is defined to be the "union of the closures of $U_j$ intersecting $K_n$".
So $U_j$ is an open set that is a subset of $K_{n+1}$ and so $U_j \subseteq K_{n+1}^\circ$ almost trivially or by definition. As this holds for all such $U_j$ it holds for their union too.
Hence the inclusion. C'est tout.