$S^2$ without $n$ points is homeomorphic to $S^2$ without $m$ points if and only if $n = m$ 
Consider the unit sphere $S^2$ with the subspace topology of $\mathbb{R}^3$. Now let $n,m$ be positive integers. Prove that $S^2$ with $n$ different points removed from it is homeomorphic to $S^2$ with $m$ different points removed from it if and only if $n = m$.

Let's assume that the $n$ points removed are $A = \{a_1, \cdots, a_n \} \subset S^2$ and that the $m$ points removed are $B = \{b_1, \cdots, b_m \} \subset S^2$.
First, assume that $n = m$. Using the stereographic projections, $S^2 \setminus A$ is homeomorphic to $\mathbb{R}^3$ without $n-1$ points and $S^2 \setminus B$ is homeomorphic to $\mathbb{R}^3$ without $m-1$ points. Since $n = m$, I assume that $\mathbb{R}^3$ without $n-1$ points is homeomorphic to $\mathbb{R}^3$ without $n-1$ (possible different) points, but I don't know how to explicetly prove this.
If $S^2 \setminus A$ is homeomorphic to $S^2 \setminus B$, then, using again the stereographic projection, if $n \neq m$, then I assume that $\mathbb{R}^3$ without $n-1$ points is not homeomorphic to $\mathbb{R}^3$ without $m-1$ points, but again, I don't know how to prove this.

Not sure why algebraic topology seems to be eschewed here, but for the harder direction I would guess that that $H_1(X)\cong\Bbb Z^{n-1}$, whereas $H_1(Y)\cong\Bbb Z^{m-1}$.
For this one can use that $X$ deformation retracts onto a "rose with $n-1$ petals". 
