On the geometrical form of the Hahn-Banach theorem

Question

I'm currently working on the geometrical form of Hahn-Banach theorem:

Hahn-Banach, first geometric form Let $A \subset E$ and $B \subset E$ be two nonempty convex subsets such that $A \cap B=\emptyset$. Assume that one of them is open. Then there exists a closed hyperplane that separates $ A$ and $B$.

In the proof, it is used that $A-B$ is convex. I wonder if we may construct two non-convex sets $A$ and $B$ such that $A-B$ is convex.

Edit: Sorry, I forget to precise that $A\cap B= \emptyset$.

Answer 1

In $\mathbf{R}^2$ it holds that $S^1-S^1=B_2(0)$