Say we have a set $S\subset\Bbb{R^n}$, and $p\in\text{int}(S), q\in\text{int}(\Bbb{R^n}\setminus S)$. I have to prove that every path between $p,q$ goes through $\partial S$ (the boundary of S). So I let $\gamma:[0,1]\rightarrow \Bbb{R^n}$ be a continuous path s.t $\gamma(0)=p, \gamma(1)=q$, and I want to prove that there exists $c\in [0,1]$ s.t $\gamma(c)\in\partial S$.
My attempt was assuming by contradiction that such $c$ doesn't exist and reach a contradiction, but I don't know how to even start. Intuitively, this statement makes sense but I don't know how to prove it.
I did notice that $\gamma ([0,1])$ is compact, though I don't know if it helps. I want to show that if there doesn't exists such $c$, then $\gamma([0,1])\subseteq\text{int}(S)$ or $\gamma([0,1])\subseteq\text{int}(\Bbb{R^n}\setminus S)$.
For every set $S$ we have that $$\Bbb R^n = \rm{int}(S) \cup \partial S \cup \rm{int}(\Bbb R^n \setminus S)$$
where the union is (pairwise) disjoint. $p$ is in the leftmost set, $q$ in the rightmost. So if $\gamma[[0,1]] \cap \partial S = \emptyset$, we could write
$$\gamma[[0,1]] = \left((\gamma[[0,1]] ) \cap \rm{int}(S)\right) \cup \left((\gamma[[0,1]] ) \cap \rm{int}(\Bbb R^n \setminus S)\right)$$
which makes the connected set $\gamma[[0,1]]$ a disjoint union of two non-empty ($p$ and $q$ are witness to that) and disjoint relatively open sets, contradiction.
So the intersection with $\partial S$ is not empty.