Let $U\subset S^n$ open subset of the sphere homeomorphic to $\mathbb{R}^n$ prove that $\mathbb{S}^n \setminus U$ is connected
Assume that $\mathbb{S}^n \setminus U$ isn´t connected then there exists $A,B\subset \mathbb{S}^n$ closed disjoint subsets such that $\mathbb{S}^{n}\setminus U=A\cup B$, since $A,B$ are closed subsets of a compact space there are compact.
Since $A,B$ are compact by Wallace theorem there are open disjoint sets $R,S$ such that $A \subset R$ and $B \subset S$.
Let $f:\mathbb{R}^n \longrightarrow U$ a homeomorphism, then notice that $\mathbb{S}^n \setminus A \cup B$ is compact in U.Therefore there exists $r>0$ such that $\mathbb{S}^n \setminus A \cup B \subset f(\overline{B(0,r)})$ .
Lat us denote $K=f(\overline{B(0,r)})$ which is compact.
We must check that $R \cap K= \emptyset$ and $S\cap K= \emptyset$.
The connected set $U\setminus K=f(\mathbb{R}^n-\overline{B(0,r))}\subset R \cup S$.
Now it suffices to show that $(U \setminus K) \cap R = U \cap R = \emptyset$, and $ (U − K) ∩ S = U ∩ S = \emptyset$ to attain an absurd with the connectedness of $\mathbb{S}^n$.
There are other ways of prove this result? I have been tryed usin poolar coordinates, and find a function $f:X \to \mathbb{S}^n \setminus U$ where $X$ is path connected to conclude that $\mathbb{S}^n \setminus U$ is also connected, but any idea comes to my mind, any suggestion or other proof diferent that the above?