proof of invariance of dimension
An application of the invariance of dimension theorem shows that ${\mathbb{R}}^{n}$ is homeomorphic^{} to ${\mathbb{R}}^{m}$ if and only if $m=n$. Already this is a difficult question. (We will assume $n\le m$ throughout this article.)
Simple arguments suffice for small dimensions^{}.

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If $n=0$ cardinality is sufficient: there can be no bijection between ${\mathbb{R}}^{0}=\{0\}$ and ${\mathbb{R}}^{n}$, $m>0$, as the latter is uncountable.

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If $n=1$, then suppose $f:\mathbb{R}\to {\mathbb{R}}^{m}$ is a homeomorphism with $m>1$. Then certainly the following restriction^{} of $f$
$$f:\mathbb{R}\{0\}\to {\mathbb{R}}^{m}\{f(0)\}$$ is also a homeomorphism. Yet, as $m>1$, ${\mathbb{R}}^{m}\{f(0)\}$ is (path) connected^{} but $\mathbb{R}\{0\}$ is not connected. Thus this restriction of $f$ cannot be a homeomorphism so indeed the original $f$ could not be a homeomorphism.
Unfortunately neither of these two arguments extends well to the cases where $n,m>1$. Indeed even the case for $n=1$ requires a reasonable amount of work to fill in the details. However, the latter approach does provide the necessary hint for a full solution.
To solve the problem outright depends on algebraic invariants from homology^{}, a surprisingly big hammer for such a basic topological question. But the conceptual steps are still basic, and we will attempt to highlight them in our exposition of the proof.
Let $U$ and $V$ be nonempty open subsets of ${\mathbb{R}}^{n}$ and ${\mathbb{R}}^{m}$ respectively. Assume that $f:U\to V$ is a homeomorphism.
Choose a point $x\in U$ (akin to the point we removed when $n=1$.) Then consider the relative homology groups ${H}_{i}(U,U\{x\})$, $i\in \mathbb{N}$. As $U$ is open we may apply the Excision Theorem (axiom) to claim ${H}_{i}(U,U\{x\})\cong {H}_{i}({\mathbb{R}}^{n},{\mathbb{R}}^{n}\{x\})$ – basically, to look at a punctured open disk it to look at a punctured ${\mathbb{R}}^{n}$. Now we look at the induced long exact sequence from the relative pair $({\mathbb{R}}^{n},{\mathbb{R}}^{n}\{x\})$ and find ${H}_{i}({\mathbb{R}}^{n},{\mathbb{R}}^{n}\{x\})$ is isomorphic^{} to the reduced homology ${\stackrel{~}{H}}_{i}({\mathbb{R}}^{n}\{x\})$. But ${\mathbb{R}}^{n}\{x\}$ contracts to the sphere ${S}^{n1}$ – and homologoy preserves homotopy type^{} – so we now have ${H}_{i}(U,U\{x\})\cong {H}_{i}({S}^{n1})$. (Puncture a disk, and it deflates to a sphere of lower dimension.)
Now it is an exercise in homology to prove that ${H}_{i}({S}^{n1})=0$ if $i\ne 0,n1$ and $\mathbb{Z}$ otherwise. In particular we are using the fact that the invariance of dimension of spheres is (more) easily established by the homology groups.
We now repeat the process with $V$. If $U$ and $V$ are indeed homeomorphic, then this process will result in isomorphic homology groups for every $i\in \mathbb{N}$. In particular,
$$\mathbb{Z}\cong {H}_{m1}({S}^{m1})\cong {H}_{m1}(V,V\{f(x)\})\cong {H}_{m1}(U,U\{x\})\cong {H}_{m1}({S}^{n1}).$$ 
Thus either $m=1$ which implies $n=0,1$ as $n\le m$, or $m=n$. If $n=0$ we have already seen $m=n$. So the result stands for all $m,n$.
For a detailed accounting of this theorem together with the necessary lemmas refer to:
Allen Hatcher, Algebraic Topology, Cambridge University Press, Cambridge, 2002. Available online at: http://www.math.cornell.edu/ hatcher/AT/ATpage.htmlhttp://www.math.cornell.edu/ hatcher/AT/ATpage.html
Title  proof of invariance of dimension 

Canonical name  ProofOfInvarianceOfDimension 
Date of creation  20130322 15:56:39 
Last modified on  20130322 15:56:39 
Owner  Algeboy (12884) 
Last modified by  Algeboy (12884) 
Numerical id  8 
Author  Algeboy (12884) 
Entry type  Proof 
Classification  msc 5500 