Is the intersection connected if union is simply connected?

No, this is not true in general (at least not without further assumptions on the two sets $A$ and $B$, such as both being open or both closed).

Imagine a solid rectangle. You can draw two L-shaped sets $A$ and $B$ in your rectangle, one of which is open, and the other closed, such that their union is the entire space, but their intersection consists of 2 disjoint rectangles.

Edit: Here is a picture of the counterexample:

Edit 2: The question has been edited so that $\mathrm{int}(A) \cup \mathrm{int}(B)$ is simply connected. In this case this example no longer works for obvious reasons, but I will leave the answer here as a counterexample to the more general situation where $A$ and $B$ can be any subsets of the space.

No this is not true. You can modify the standard example of two half circles with "half circles" of the topologist's circle. This ensures that the union is simply connected but you get the same result as with the usual half circles.