A contractible open subset of $\mathbb{R}^n$ need not be homeomorphic to $\mathbb{R}^n$. The Whitehead manifold is an open subset of $\mathbb{R}^3$ which is contractible but not homeomorphic to $\mathbb{R}^3$ (it is not simply connected at infinity).
Removing a point from $\mathbb{R}^3$, we obtain a space homotopy equivalent to $S^2$. Is this true of any open contractible subset of $\mathbb{R}^3$? In particular:
Is Whitehead's manifold with a point removed homotopy equivalent to $S^2$?
Yes. Deleting a point from a manifold of dimension $n \geq 3$ doesn't change the fundamental group, so the result is still simply connected. (Either use van Kampen or transversality.) Mayer-Vietoris shows that the homology of the resulting space is the same as $S^2$, whence $\pi_2(U \setminus p)$ is isomorphic to $H_2(U \setminus p) = \Bbb Z$, generated by a small sphere around $p$. Because the map $S^2 \to U \setminus p$ is an isomorphism on homology, and $U \setminus p$ is simply connected, Whitehead + repeated application of Hurewicz shows that the map is a homotopy equivalence. This works in any dimension $n \geq 3$.
(Of course, it's also true in dimension 2, where the only open contractible surface is homeomorphic to $\Bbb R^2$, as well as dimension 1, where the only open curve is $\Bbb R$. Indeed the same proof applies in the first case, where van Kampen instead implies that the result when you delete a point has fundamental group $\Bbb Z$.)
Of course, it is not proper homotopy equivalent to $\Bbb R^3 \setminus 0$, as proper homotopy equivalences preserve the fundamental groups at infinity of the ends.