I do not expect any simple proofs of this result. A good exercise would be to prove this for domains in $R^2$ without using anything about the classification of surfaces or the Schoenflies theorem in $R^2$.
Here is a proof of your statement (in the topological category). I will assume that $n\ge 2$ since there is nothing to prove in dimension 1.
Let $U$ be any open (connected, which is not really necessary) domain in $R^n$ whose 1-point compactification is an $n$-manifold $M$. First of all, it is easy to see that the complement of $U$ has to be connected (since a point cannot locally separate a manifold of dimension $\ge 2$).
Let $p\in M$ be such that $M-p\cong U$. Let $B$ be a metric ball centered at $p$ (with respect to a Euclidean metric on a neighborhood of $p$ in $M$). Let $Y=\partial B$. Then $Y$ as an $n-1$-dimensional tame sphere
is contained in $U$ and separating $R^n$ in two components, a bounded component $C$ and an unbounded one. The bounded component is necessarily contained in $U$ (since the complement of $U$ is connected). Then, by topological Schoenflies theorem, $C$ is homeomorphic to $B^n$. Thus, $M$ is obtained by gluing two balls ($B$ and $C$) along their common boundary sphere $Y$ and, hence, homeomorphic to $S^n$.