Proof:

The first part follows from the fact that the homology groups $H_n(S^m)$ and $H_n(S^n)$ are non-isomorphic. If $\mathbb{R}^m$ and $\mathbb{R}^m$ were homeomorphic then their one-point compactifications would also be homeomorphic which means $S^n$ and $S^m$ would be homeomorphic leading to a contradiction.