Degree theory and Invariance of domain We'll use the Proposition (F) to show that: 
(Invariance of domain) Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected.
If $f$ is injective then $f$ is a open map.
dem: We can assume that $f$ is locally proper by $f:M→Y$ $C^{1}$ map between manifolds $M$ e $N$ with dimension m and n respectively then $f$ is locally proper.
Let $x \in M$, and $y:=f(x) \in N$ with $(U,\phi)$ chart in $M$ of $x$ and $(V,\psi)$ chart in $N$ of $f(x)$ such that $f(U)\subseteq V$ is proper. Let $B(\psi(y), \epsilon) \subseteq \psi(V)$ 
so  $\psi^{-1}(B(\psi(y), \epsilon)$ is a open  connected subset of $N$. Since $f$ is continous, exist $W$ open in $M$ such that $W \subseteq U$ and $f(W) \subseteq \psi^{-1}(B(\phi(x), \epsilon)$.
Let  $B(\phi(x), \delta) \subseteq \phi(W)$ so $\phi^{-1}(B(\phi(x), \delta)$ is a open  connected subset of $M$ such that 
$f(\phi^{-1}(B(\phi(x), \delta)) \subseteq  \psi^{-1}(B(\psi(y), \epsilon)$.
Note that $\tilde{f} = f|_{\phi^{-1}(B(\phi(x), \delta)}: \phi^{-1}(B(\phi(x), \delta) \to  \psi^{-1}(B(\psi(y), \epsilon)$  is injective proper mapping of two oriented, boundaryless, smooth manifolds of dimension $m$, where 
where $\psi^{-1}(B(\psi(y), \epsilon)$ is connected. Then by (F) $\tilde{f}$ is surjective. So $f( \phi^{-1}(B(\phi(x), \delta))$ is open, and we finished.
In this solution has a problem: I'd take $\psi^{-1}(B(\psi(y), \epsilon + \xi)$  and the result continous true, so $\tilde{f}$  is a bijecion of $\phi^{-1}(B(\phi(x), \delta)$ to  $\psi^{-1}(B(\psi(y), \epsilon)$ and $\psi^{-1}(B(\psi(y), \epsilon +\xi)$, and it's false. 
Where am I missing??? 
Thank you.

We have that:
Definition (A):
Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; and $x$ is a regular point of $f$.
So $df_{x}$ is a linear isomorphism. If  $f$ preserves (resp. reverse) orientation at $x$  we denote $sing_{x}(f) = +1$ (resp. $-1$).
(Mapping Degree, Ruiz, 2009. p. 86)
Proposition (B):
Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; and $y$ is a regular value of $f$.
so $f^{-1}(y)$ is a finite set and $d(f,y)= \sum_{x \in f^{-1}(y)} sign(df_{x})$.
(Mapping Degree, Ruiz, 2009. p. 95)
Proposition (C):
Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected.
Then the integer $d(f,y)$ does not depend on the choice of regular valeu $y$. It is called the degree of $f$ an is denoted by $deg(f)$.
(Mapping Degree, Ruiz, 2009. p. 99)
Corollary (D):
Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected.
If the $deg(f) \neq 0$ then $f$ is surjective  map.
(Mapping Degree, Ruiz, 2009. p. 99)
Proposition (E): 
If $f: U \to \mathbb{R}^{n}$ $C^{1}$ locally injective and $U$ is a open set in $\mathbb{R}^{n}$ then the set of the points $x \in U$
such that $df_{x}$ is injective is open  and dense in $U$.
dem: By $rank$ $theorem$.
Proposition (F):
Let $f: M \to N$ $C^{1}$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected.
If $f$ is injective then  $f$ is surjective.
dem: Let $(U,\phi)$ chart in $M$. By (E) there exists $x \in U$ such that $df_{x}$ is injective, so $df_{x}$ is a isomorphism linear. So $y=f(x)$ is regular value, 
and since $f$ is injective we have that $d(f,y) \neq 0$. So by (D), we have that $f$ is surjective.
 A: Your proof of invariance of domain doesn't work, because you have given no justification that $\tilde{f}$ is proper.  In fact, it probably won't be proper.  Indeed, if (as you suggest) you replace $\epsilon$ by $\epsilon+\xi$ for some $\xi>0$, then $\tilde{f}$ will definitely not be proper, since $\psi^{-1}(\overline{B(\psi(y), \epsilon)})$ is a compact subset of $\psi^{-1}(B(\psi(y), \epsilon+\xi))$ containing the image of $\tilde{f}$, but the domain of $\tilde{f}$ is not compact.
To fix the proof, you need to be much more careful in choosing the domain of $\tilde{f}$.  You can take the domain of $\tilde{f}$ to be $f^{-1}(\psi^{-1}(B(\psi(y), \epsilon)))$: then $\tilde{f}$ will be proper because $f$ was.  (Properness of a map is preserved if you restrict the range to a subset $A$ and the domain to $f^{-1}(A)$.)  To conclude that $\tilde{f}$ (and hence $f$) is open, you then need to use the fact that a proper continuous bijection between locally compact Hausdorff spaces is a homeomorphism.
