Can this be a way to prove that $\Bbb{R}^2$ and $\Bbb{R}^3$ are not homeomorphic? The normal way I use to prove that $\Bbb{R}$ and $\Bbb{R}^2$ are not homeomorphic is by removing a point and then using path connectedness.
But this method doesn't seem to work for $\Bbb{R}^m$ and $\Bbb{R}^n$ and it ends up that its better if you use fundamental groups.
But what if I remove a line instead of a point from both the topological spaces. Does the argument still fail or will it lead to a proof?
 A: If $m<n$, then we can make $\mathbb{R}^m$ disconnected by deleting an $(m-1)$-dimensional subset; more snappily, a codimension 1 subset. Now, in order to show that $\mathbb{R}^n\not\cong\mathbb{R}^m$, it will suffice to show that, if we delete the homeomorphic image of $\mathbb{R}^{m-1}$ from $\mathbb{R}^n$, the latter remains connected.
This is where things get tricky. For $m=1$ this is easy, since the homeomorphic image of a point is . . . well, a point. :P But the possible homeomorphic images of even a line in $\mathbb{R}^n$ are suddenly extremely complicated! I suspect that patching this argument will require using fundamental groups.

EDIT: In the comments, the OP writes: "I  want to see for a line which might be a little wacky." Well, first of all, the point is not that there are (homeomorphic images of) lines (in $\mathbb{R}^n$) which are wacky, but rather that it would take proof to show there aren't any. As evidence that you should not take this for granted:


*

*The Jordan curve theorem (https://en.wikipedia.org/wiki/Jordan_curve_theorem) is hard.

*This isn't a line, but it's pretty darn cool, and it should shake your intuitions about lines:  https://en.wikipedia.org/wiki/Pseudo-arc.

*Not a homeomorphic image, but space-filling curves (https://en.wikipedia.org/wiki/Space-filling_curve) should cause some worry. In general, fractals probably ought to give one pause.

*Generally, I think algebraic and low-dimensional topology and algebraic geometry will have a lot to say about how weird codimension-one subspaces can be!

FURTHER EDIT: On the other hand, this doesn't rule out the possibility of arguments avoiding algebraic topology. For a really neat example, check out http://arxiv.org/pdf/1003.1467v2.pdf, which uses set theory of all things to show $\mathbb{R}^2\not\cong\mathbb{R}^n$ for any $n>2$. As far as I can tell, however, their method does not generalize beyond 2.
And see also Elementary proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$.
A: Essentially, yes, but it is quite hard to prove that removing a subspace homeomorphic with $\mathbb{R}$ (your "line") cannot disconnect $\mathbb{R}^3$. The Jordan curve theorem states that any continuous embedding of a circle in the plane disconnects it and that implies that $\mathbb{R}^2$ and $\mathbb{R}^3$ are not homeomorphic via an argument along the lines of the one you have in mind. Brouwer's theorem on the invariance of domain generalises this and implies that $\mathbb{R}^m$ and $\mathbb{R}^n$ are homeomorphic iff $m = n$. Both the Jordan curve theorem and invariance of domain are usually proved using homology theory.
A: The tricky and counterintuitive fact is that there is a surjective continuous map $g:\mathbb{R}\rightarrow \mathbb{R}^2$(called space-filling curve).
So, by producting with $id:\mathbb{R}\rightarrow \mathbb{R}$, you have a surjective continuous map $f:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ which maps $\mathbb{R}\times \left\{0\right\}$ to $\mathbb{R}^2 \times \left\{0\right\}$.
So, in this case, connectedness argument fails.
(Because there is no homeomorphism between $\mathbb{R}^2 $and $\mathbb{R}^3$, such argument does not cause contradiction, but at least this shows that using connectedness is not a smartest way to show the two space are not homeomorphic)
