Is a subset of $\mathbb R^2$ homeomorphic to an open subset of $\mathbb R^2$ an open subset? Suppose $X\subset \mathbb R^2$ with the subset topology and $U\subset\mathbb R^2$ an open subset with the subset topology, if $X\cong U$ and we deduce $X\subset \mathbb R^2$ is also open? 
This is not true for general topological space, for example let $S=\{1,2\}$ with the topology
\begin{equation}
 \tau_S=\big\{\emptyset,\{1,2\},\{1\}\big\}
\end{equation}
then 
\begin{equation}
 \{2\}\cong\{1\}
\end{equation}
but $\{1\}\subset S$ is open and $\{2\}\subset S$ is not open. 

Yes, this is true for any $\mathbb{R}^n$ as a consequence of the invariance of domain theorem. If
$$f:U\to X$$
is a homeomorphism then we have an injective continuous map $g:U\to\mathbb{R}^n$, $g(x)=f(x)$ which has to be open. Thus the image, being $X$ is open.
