this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof (which is, quoting from this post)
A linear map $\Bbb R^n \to \Bbb R^n$ can be understood to preserve or reverse orientation, depending on whether its determinant is $+1$ or $-1$. This notion of orientation can be generalized to arbitrary homeomorphisms, giving a "degree" $\deg(m)$ for every homeomorphism which is $+1$ if it is orientation-preserving and $-1$ if it is orientation-reversing. The generalization has all the properties that one would hope for. In particular, it coincides with the corresponding notions for linear maps and differentiable maps, and it is multiplicative: $\deg(f \circ g) = \deg(f)\cdot \deg(g)$ for all homeomorphisms $f$ and $g$. In particular (fact 1), if $h$ is any homeomorphism whatever, then $h \circ h$ is an orientation-preserving map.
Now, suppose that $h : X^2 \to \Bbb R^3$ is a homeomorphism. Then $X^4$ is homeomorphic to $\Bbb R^6$, and we can view quadruples $(a,b,c,d)$ of elements of $X$ as equivalent to sextuples $(p,q,r,s,t,u)$ of elements of $\Bbb R$.
Consider the map $s$ on $X^4$ which takes $(a,b,c,d) \to (d,a,b,c)$. Then $s \circ s$ is the map $(a,b,c,d) \to (c,d,a,b)$. By fact 1 above, $s \circ s$ must be an orientation-preserving map. But translated to the putatively homeomorphic space $\Bbb R^6$, the map $(a,b,c,d) \to (c,d,a,b)$ is just the linear map on $\Bbb R^6$ that takes $(p,q,r,s,t,u) \to (s,t,u,p,q,r)$. This map is orientation-reversing, because its determinant is $-1$. This is a contradiction. So $X^4$ must not be homeomorphic to $\Bbb R^6$, and $X^2$ therefore not homeomorphic to $\Bbb R^3$.
and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.