m-manifold of $\mathbb{R}^m$ is open I'm trying to prove the following:
Show that a subset $M \subset \mathbb{R}^m$ is an $m$-dimensional manifold if and only if $M$ is open.
Where for $m$-dimensional manifold one uses the definition of Milnor's "Topology from the differentiable viewpoint" 
i.e. $\forall p\in M  \exists  $ $U\subset \mathbb{R}^m$ open neighbourhood of p and $\Omega \subset \mathbb{R}^m$ open and a diffeo.  $ \varphi:M\cap U\to\Omega$.
I want to show that $\varphi^{-1}$ is an open map $\Omega\to \mathbb{R}^m$  but I can't find how, can someone give me some hints ? 
 A: I'll prove that if $M$ is a manifold of dimension $m$ on $\mathbb R^m$ then $M$ is a open set of $\mathbb R^m$. The other part, that all open subsets of $\mathbb R^m$ is manifold is trivial.
Take $x\in M$ and lets prove that $x$ is a interior point of $M$. There is a open neighborhood $U$ of $x$ on $M$ and a diffeomorphism $\phi:U \to V$, where $V$ is a open set of $\mathbb R^m$. Just to be more precise, $\phi$ is a diffeomorphism in the sense of Milnor's book, that is, there are an open subset $U'$ of $\mathbb R^m$ such that $U = U'\cap M$ and a diffeomorphism (in the usual sense) $\Phi: U' \to V$ such that $\Phi|_U = \phi$.
Notice $\psi = \phi^{-1}$ is a smooth map from $V$ to $\mathbb R^m$ and $d\psi$ is invertible on all points of $V$. By the inverse function theorem there is a neighborhood $W$ of $y=\phi(x)$ such that $\phi(W)$ is open on $\mathbb R^m$, $W\subset V$ and $\psi:W \to \psi(W)$ is diffeomorphism. Since $\psi(W)$ is an open neighborhood of $x$ and $\psi(W)\subset U \subset M$ we conclude that $x$ is interior point of $M$. Therefore $M$ is open on $\mathbb R^m$.
