Elementary proof of topological invariance of dimension using Brouwer's fixed point and invariance of domain theorems? http://people.math.sc.edu/howard/Notes/brouwer.pdf https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/
These papers give fairly elementary proofs of Brouwer's fixed point and invariance of domain theorems. Having established these tools, is it possible to prove that an open subset of $ \mathbb{R}^n $ and an open subset of $ \mathbb{R}^m $ can't be homeomorphic unless $ n = m $ without refering to all those more advanced things which are normally used here, such as homology theory and algebraic topology?

If $V$, open subset of $\Bbb R^m$, were homeomorphic to an open subset of $\Bbb R^n$, $U$, let $f: U \to V$ be a homeomorphism. (Suppose WLOG that $m \leq n$.) Compose with a linear inclusion map $\Bbb R^m \hookrightarrow \Bbb R^n$ to get a continuous injective map $U \to \Bbb R^n$ with image contained in the subspace $\Bbb R^m$, as it factors as $U \to V \subset \Bbb R^m \hookrightarrow \Bbb R^n$. 
If $m < n$, the image cannot be open - any neighborhood of a point in the hyperplane contains points not in the hyperplane. Therefore $m=n$. So invariance of domain implies invariance of dimension.
Of course, this says even more: there's not even a continuous injection $U \to V$ between open sets of $\Bbb R^n, \Bbb R^m$ respectively, $n>m$.
