Any open set in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ I believe the claim is right. But I can only prove for the case where the open set is convex.
EDIT: As someone pointed out, it is not true in general. How about "simply connected open set"?
 A: A set of conditions for which this becomes true is given on another question, but the proof appears to be quite nontrivial. In brief, an open set which is contractable and satisfies a certain "boundary" connectedness condition is homeomorphic to $\Bbb R^n$.
To address the edit: contractibility is a stronger condition than simply connectedness, and so we would not necessarily expect it to be enough. Indeed it is not. Take an open annulus and spin it around a diameter in 3-space; the result is a "thickened" 2-sphere. This is an open set which is not homeomorphic to $\Bbb R^3$.
In some sense, simply connectedness removes the possibility of "1-dimensional holes" (i.e. holes that can be surrounded by a 1-sphere). But you need to also account for higher dimensional holes in general; this is the intuitive notion formalized by contractibility.
A: It is not true in general. For example, $\mathbb R^2$ minus one point is open but not homeomorphic to $\mathbb R^2$ (say, because it isn't simply connected).
A: In general the result is not true. For example $B(0,1)\cup B((2,\ldots,0),\frac{1}{2}))$ is an open set and it cannot be homeomorphic to $\mathbb{R}^n$ because $\mathbb{R}^n$ is connected but the topological space  $B(0,1)\cup B((2,\ldots,0),\frac{1}{2}))$ is not connected.
