Show that $\mathbb{R}^m$ is not homeomorphic to $\mathbb{R}^n$ if $m\ne n$. You may assume that $S^m$ and $S^n$ are different homotopy type if $m\ne n$.
My attempt: Suppose $\mathbb{R}^m$ is homeomorphic to $\mathbb{R}^n$. Since $\mathbb{R}^m$ is homeomorphic to $A^m=S^m-\{x\}$, the sphere minus a point, then we would have $A^m$ is homeomorphic to $A^n$. Then $A^m$ and $A^n$ have the same homotopy type. Then by definition, there are functions $f:A^m\to A^n$ and $g:A^n \to A^m$ such that $g\circ f=id_m$ and $g\circ f=id_n$.
But then I don't know how to related that to $S^m$ and $S^n$? I know that $S^n$ and $S^n-\{x\}$ are definitely not the same homotopy type though...
Assume that $n<m$. If $\mathbb{R}^{n}\approx \mathbb{R}^{m}$ then $$S^{n-1}\cong \mathbb{R}^{n}\setminus \{x\}\approx \mathbb{R}^{m}\setminus \{y\}\cong S^{m-1},$$ where $x$ is any point in $\mathbb{R}^{n}$ and $y$ is the image of $x$ under the homeomorphism $\mathbb{R}^{n}\approx \mathbb{R}^{m}$, and $\cong$ means homotopy equivalent. This gives a contradiction with your assumption about the homotopy types. Or alternatively, you can make a direct argument as follows. If $n=1$ then $S^{n-1}$ is a two point set and thus disconnected, and $S^{m-1}$ is connected, a contradiction. So we may assume that $n\geq 2$. Now if $i<k$ then $\pi_{i}(S^{k})=0$. Hence $$\mathbb{Z}=\pi_{n-1}(S^{n-1})=\pi_{n-1}(S^{m-1})=0,$$ yet another contradiction. The case $m<n$ is similar. So we must have $\mathbb{R}^{n}\not\approx \mathbb{R}^{m}$ when $n\neq m$.