Interior of a contractible set

Question

Let $O, F \subset \mathbb R^n$ for $n \ge 2$. Suppose $O$ is open, $F$ is closed and $O \subset F$ with closure $\bar{O} = F$. If we further assume that both $O$ and $F$ are contractible to the same fixed point $p \in O$. Can we conclude that $F^{\circ} = O$?

Answer 1

No. Let $F = D^n = \{ x \in \mathbb{R}^n \mid \lVert x \rVert \le 1 \}$ and $O = \mathring{F} \setminus [1/2,1) \times \{ (0,\dots,0) \}$. Both are contractible to $0$.