How do I prove that $S^n$ is homeomorphic to $S^m \Rightarrow m=n$? This is what I have so far:
Assume $S^n$ is homeomorphic to $S^m$. Also, assume $m≠n$. So, let $m>n$.
From here I am not sure what is implied. Of course in this problem $S^k$ is defined as:
$S^k=\lbrace (x_0,x_1,⋯,x_{k+1}):x_0^2+x_1^2+⋯+x_{k+1}^2=1 \rbrace$ with subspace topology.

A non-elementary argument would be that spaces with non-identical homotopy groups are not homeomorphic. 
Assume $n < m$. Then $\pi_n (S^n) = \mathbb Z$ (see this article) but $\pi_n(S^m) = 0$. Hence $S^n$ and $S^m$ are not homeomorphic if $n \neq m$.
Alternatively, as pointed out in the comments, you could use homology groups:
$H_n(S^n) = \mathbb Z$ but $H_n(S^m) = 0$. But homeomorphic spaces have isomorphic homology groups.
