$\mathbb{R}^n$ without finitely many points are homeomorphic I am kind of stuck when dealing with the following question:

Let $A,B \subset \mathbb{R}^{n}$ be two finite sets with $|A|=|B|$. Prove that $\mathbb{R}^{n}\setminus A $ is homeomorphic to $\mathbb{R}^{n}\setminus B$.

My attempt is to seek a continuous bijection between these two metric spaces and try to show that the inverse function is also continuous. However, I am stuck in the first step. 
Thanks.
 A: You may assume that the $x$-coordinates of all the points in a given set ($A$ or $B$) are different $-$ if they are not, just choose a different direction as your $x$-axis. So now you can order the points, and match each point $a_i\in A$ with a point $b_i\in B$. Now your bijection will use an 
appropriate shear transformation to map the horizontal region between $a_i$ and $a_{i+1}$ to the corresponding horizontal region between $b_i$ and $b_{i+1}$, with $a_i$ and $a_{i+1}$ mapping to $b_i$ and $b_{i+1}$ respectively.
I hope this is clear. If not, let me know.
A: In my comment, it doesn't mean you need to use induction, in fact, it's useless here. I think you can find that when you take those point off, you're actually restricting the map to focus on those points and map them to the range continuously and then take them off. If you found that, the following process would be quite obvious. And I'm sorry for replying so late. 

Let $A=\{x_1,x_2,...,x_k\}$ those points need to be removed, but let's don't remove it now. and $B=\{y_1,y_2,...y_k\}$ the range set.
First, define a continuous map $f:\Bbb{R}^n\to\Bbb{R}^n$ such that $f(A)=B$ (i.e. we map the point $x_i$ to $y_i$). We set $f$ to be a homeomorphism $\implies$ $f$ is a continuous bijection with continuous inverse. We can set that because there exists a whole bunch of continuous bijections that satisfy our criterion.
Define $\bar{f}:\Bbb{R}^n\setminus A\to \Bbb{R}^n,\bar{f}(x)=f(x)$ then $\bar{f}$ is continuous because it's the composite of the inclusion map $i:\Bbb{R}^n\setminus A\to \Bbb{R}^n$ and $f$, both of which are continuous.
A similar argument shows that $f^{-1}|_{\Bbb{R}^n\setminus f(A)}$ is also continuous.
Combine the information above, we get the restriction map $\bar{f},\bar{f}^{-1}$ which are continuous and bijective (directly followed by $f$ is bijective).
Thus $\Bbb{R}^n\setminus A\cong\Bbb{R}^n\setminus B,$ if $|A|=|B|<\infty$.

Extension:
Can you manage to show that if $g:X\to Y$ is a homeomorphism, and $A\subset X$ being given the subspace topology, then $X\setminus A\cong Y\setminus g(A)$? Using exactly the same method mentioned above.
