proving that $\mathbb{R}^{2}$ and $\mathbb{R}^{n}$ are not homeomorphic I'm struggling for proving that $\mathbb{R}^{2}$ and $\mathbb{R}^{n}$ are not homeomorphic for $n \ge3.$ My approach is similar to proof of the state below.

$\mathbb{R}$ and $\mathbb{R}^{n}$ are not homeomorphic for n$\ge$2.

More specifically, suppose to the contrary that there is homeomorphism $h:\mathbb{R}^{2}\rightarrow \mathbb{R}^{n}$ . Since $\mathbb{R}^{2}-\mathbb{S}^{1}$ is NOT connected and it has separation $\{x\in \mathbb{R}^{2} \mid \left|x \right|<1\}$ and $\{x\in \mathbb{R}^{2} \mid \left| x \right| >1\}.$ I know intuitively that $\mathbb{R}^{n}-h \left( \mathbb{S}^{1}\right)$ is connected, but I cannot prove mathematically.
 A: Your approach works smoother if you flip the direction around.
Suppose you have a homeomorphism $f:\mathbb R^n\to \mathbb R^2$ and consider a fixed closed nonintersecting curve $\gamma$ in $\mathbb R^n$ that you can describe nicely. If you choose a nice straightforward circle, it will be easy to show that $\mathbb R^n\setminus\gamma$ is path-connected. On the other hand, $f(\gamma)$ is a Jordan curve in $\mathbb R^2$, so $\mathbb R^2\setminus f(\gamma)$ is not connected, a contradiction. (This assumes you can appeal to the Jordan curve theorem without proving it yourself).
A: Since you tagged this with algebraic topology, I will give you a very simple argument for this. Let $1\le m<n$ and suppose there was a homeomorphism $f\colon\mathbb{R}^m\to\mathbb{R}^n$. Removing any $x\in\mathbb{R}^m$ and the corresponding point $f(x)$ from their respective spaces, we have (still homeomorphic) punctured euclidean spaces. The first is homotopy equivalent to $S^{m-1}$, and the second to $S^{n-1}$. This gives us their homologies, $H_*(\mathbb{R}^m\setminus\{ x\}) \cong H_*(S^{m-1})$ and $H_*(\mathbb{R}^n\setminus\{ f(x)\})\cong H_*(S^{n-1})$.
We know that $n\ge2$, so $H_k(S^{n-1}) = 0$ if $k\neq 0, n-1$ and $H_0(S^{n-1})\cong \mathbb{Z} \cong H_{n-1}(S^{n-1})$. The same holds for $S^{m-1}$ (with $n$ replaced by $m$) if $m\ge2$, but if $m=1$, then $H_0(S^{m-1})\cong \mathbb{Z}\oplus\mathbb{Z}, H_k(S^{n-1})=0$ for $k\ge1$. Since the punctured spaces are homeomorphic to one another and homotopy equivalent to those spheres, the spheres are then homotopy equivalent to one another. This means that they have isomorphic homologies, but then $\mathbb{Z} \cong H_{n-1}(S^{n-1}) \cong H_{n-1}(S^{m-1}) = 0$, since $m\neq n$. This is an absurdity, and therefore the punctured spaces cannot be homeomorphic. Since the homeomorphism between them was obtained from a homeomorphism between the original euclidean spaces, such a homeomorphism cannot exist.
