Open subset of Euclidean space can't have non-spherical manifold compactification. The title is not complete, since it would be too long. Consider the following statement:

Let $U \subset \mathbb{R}^n$ be open, connected and such that its one-point compactification is a manifold. Then, this compactification must be (homeomorphic to) the sphere $S^n$.

Is the statement above true? If so, why?
 A: 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 $S^n$. 
A: I can imagine an ellementary approach only for  the special case of $\mathbb{R^2}  $ and $\mathbb{R}$.
For $\mathbb{R^2}$:We know that all the compact surfaces arise from adding to the sphere a finite amount of handles or Mobius-strips. In any case, if you remove a point from a compact surface which is something of the above but not a sphere then you wouldn't get something homemoprhic to $\mathbb{R^2}$. So the only compactification could be to a sphere.
Similar arguments goes for $\mathbb{R}$ since the only compact $1$ manifolds are a closed line segment and the circle. 
