Prove that for every $n \geq 2$, $\Bbb R$ is not homeomorphic to $\Bbb R^n$ So for them to not be homeomorphic the function or inverse of the function must not be continuous. Correct? 
Should I assume homeomorphism first, and create open balls?
 A: HINT: If you remove a point from $\Bbb R$, what’s left is not connected. If $n\ge 2$, does removing a point from $\Bbb R^n$ leave you with a connected or a disconnected set? (I think that we’ve had this before, but if so, I can’t immediately find it.)
A: This is a formal proof, with heavy tools if you know homology.
If $\mathbb R^n $ and $\mathbb R^m$ are homeomorphic thru, say,$ h$ , then the restriction of $h$ to $\mathbb R^n-{(0^n)}$ to $\mathbb R^m- {(0^m)} $ (compose with a new homeomorphism so that $0^n$ is mapped to $0^n $ if necessary) is also a homeomorphism. But $\mathbb R^n-{0^n}$ is homotopic to $S^{n-1}$  and $\mathbb R^m -{0^m} $ is homotopically-equivalent to $S^{m-1}$ ; and any two homeomorphic spaces are homotopically-equivalent.
For a partial argument for why $\mathbb R^m $ is not homeomorphic to $\mathbb R^n$, 
here is an argument for why $\mathbb R^{2n+1}$ cannot be homeomorphic to $\mathbb R^{2m}$,where maybe the "guts"of the proof can be seen, we can generalize the argument here:
http://blog.plover.com/math/R3-root.html , that uses degree theory (degree of a map, which must be +/- 1 for a homeomorphism, and degree has "nice"properties under composition.) which shows that $\mathbb R^3 $ is not ä square root", meaning $\mathbb R^3$ is not homeomorphic to the product of a space with itself. This proof can be generalized to any $\mathbb R^{2n+1}$ not being a product of a space with itself, while we have the trivial result that $\mathbb R^{2m}$ is homeomorphic to the product $\mathbb R^m \times \mathbb R^m $ . This shows $\mathbb R^{2n+1}$ is not homeomorphic to $\mathbb R^{2m}$ 
A: The classical argument to prove that $\mathbb{R}$ and $\mathbb{R}^n$ ($n \geq 2$) are not homeomorphic (given by Brian M. Scott) is really simple, elementary, and it uses a key topological property of the real line: every point is a strong cut point. The discussion Topological Characterisation of the real line on MO can be mentionned as a justification.
By the way, another argument may be:


*

*Any connected subset of $\mathbb{R}$ is arc-connected (since any such subset is an interval).

*For $n \geq 2$, $\mathbb{R}^n$ contains non-arc-connected connected subsets. (See for example Topologist's sine curve.)

A: Here's a quick and elementary proof:
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a continuous function. Take two distinct points $a,\,b \in \mathbb{R}^n$, and take two disjoint (other than at $a$ and $b$) continuous curves connecting them. The intermediate value theorem says that on each of these curves $f$ takes on every value in $[f(a),f(b)]$. But the curves are disjoint, so $f$ is not injective. So there is no such homeomorphism.
