I have a big problem understanding the following theorem
Theorem Let M, N be k-manifolds, and $f:M\to N$ an embedding. Then $f(M)$ is a submanifol of $N$.
I have seen the proof in severals books, however it is not clear what they did. So, I will write my attemp.
Firts, I will write some definitions
Sub Manifold Let $X$ be a manifold of dimension n. Let $A\subset X$. We say that $A$ is a submanifold if $\forall x\in M$, $\exists U_x \subset X$, $V_0\subset \mathbb{R}^n$ and a diffeomorphism $g:U_x \to V_0$ such that $g(U_x\cap M)=V_0\cap (\mathbb{R}\cap\times \{0\})$.
In an equivalent way, we could say that $A$ is a submanifold if there is a chart $(U,\psi)$ such that $\psi(A\cap X)$ is a sub manifold of $\mathbb{R}^n$.
Now, with these definitions I will try to prove my theorem: draft of the prove
Let $y\in f(M)$. Let $U_y\subset N$ be a open neighborhood and $(U,\psi)$ be a chart of $y$. We have that there exist $x\in M$ such that $f(x)=y$. Also, we consider $(V, \phi)$ a chart of $x$.
As $f$ is an homeomorphism, we have that $f(V)$ is open respect to the induced topology of $f(M)$, so we could have \begin{equation} f(V)=f(M)\cap U. \end{equation}
So if instead of $f$ we take $\psi^{-1} \circ f \circ \phi$ \begin{equation} \psi^{-1} \circ f \circ \phi(V)=\psi^{-1} \circ f \circ \phi (M)\cap U. \end{equation}
can I assure that?, What I am missing, and I read that with the final line we can conclude the proof but I don't know how. Thanks for your time.
Edit
I think this is the answer to my question Two definitions of embedded submanifold