An open subset $A \subseteq \mathbb{R}^{2}$ is not homeomorphic to an open subset of $\mathbb{R^{n}}$ for $n\geq 3$.
This question is from my General Topology class. We have seen up to part of algebraic topology.
My attempt:
Suppose a homeomorphism $f \colon X \to A$, where $X\subseteq \mathbb{R}^{n}$ is an open set. Then we can restrict this homeomorphism to an open ball contained in $X$ and, using some more homeomorphisms, suppose $X = \mathbb{R}^{n}$. Therefore we can also suppose $A \subseteq \mathbb{R}^{2}$ is an open simply connected set. Now, because of $\mathbb{R}^{n}-\{0\}$ is simply connected whenever $n\geq 3$, $f(\mathbb{R}^{n}-\{0\}) = A - \{f(0)\}$ is simply connected.
We want to deny that $A - \{f(0)\}$ is simply connected. In order to do that pick a $\delta > 0$ such that $B_{f(0)}[\delta] \subset A$. Then define the following function:
$$ \alpha \colon \mathbb{S}^{1} \to A - \{f(0)\}; ~ \alpha(z) = z\delta + f(0)$$
A set $X \subseteq \mathbb{R}^{n}$ is simply connected when every continuous function $p \colon \mathbb{S}^{1}\to X$ has a continuous extension $\overline{p}\colon \mathbb{B}^{2} \to X$.
I was not able to show that this function has not continuous extension to $\mathbb{B}^{2}$. Is it true? Any help would be appreciated but I would prefer help in my attempt.
Thank you.