Let $\Omega \subset \mathbb R^n$ an open and $\varphi:\Omega \to \mathbb R^n$ a local homeomorphism. Prove that $\varphi$ is an open application, that mean that $\varphi(U)$ is open for all $U\subset \Omega $.
Here's my proof. For all $x\in \Omega $, there is $U_x$ open s.t. $\varphi|_{U_x}: U_x\longrightarrow \varphi(U_x)$ is a homeomorphism. Then, $$\Omega =\bigcup_{x\in \Omega }U_x,$$ and thus $$\varphi(\Omega )=\bigcup_{x\in \Omega }\varphi(U_x)$$ which is open. Therefore $\varphi$ is open.
Does it work ?
Just to fix thing, I will add an answer. Let $A,B\subset X$. Then $\varphi: A\longrightarrow B$ is a local homeomorphism if for all $x\in A$, there is an open $U$ of $X$ that contain $x$ and an open $W$ in $X$ that contain $\varphi(x)$ s.t. $$\varphi: A\cap U\to B\cap W$$ is a homeomorphism. Now, $\varphi$ will be open mean that if you take an open $O$ of $A\cap U$ (and not of $X$), then $\varphi(O)$ will be open in $B\cap W$ (but not in $X$). In other word, for all open $O'$ of $X$, there is an open $U'$ of $X$ s.t. $$\varphi(O'\cap A\cap U)=W\cap B\cap U'.$$
Now to apply to you problem : $\Omega $ is already open in $\mathbb R^n$. It's a local homeomorphism, therefore, for all $x\in \Omega $ there is $B_x$ open in $\mathbb R^n$ that contain $x$ and $B_{\varphi(x)}$ open of $\mathbb R^n$ that contain $\varphi(x)$ s.t. $$\varphi|_{B_x\cap \Omega }: B_x\cap \Omega \longrightarrow B_{\varphi(x)}\cap \mathbb R^n=B_{\varphi(x)}$$ is a homeomorphism. Therefore, $$\varphi(\Omega) =\varphi\left(\bigcup_{x\in \Omega }B_x\cap \Omega \right)=\bigcup_{x\in \Omega }\varphi(B_x\cap \Omega )=\bigcup_{x\in \Omega }B_{\varphi(x)},$$ which is open in $\mathbb R^n$. Now we are done.