$f$ have finitely many critical points in $\Omega$

Question

Assume that $\Omega$ is an bounded open set in $R^m, f\in C^2(\overline{\Omega},R^m)$. If $f$ does not have any critical point in $\partial \Omega$, and all the critical points of $f$ in $\Omega$ are non-degenerate, prove that $f$ has finitely many critical points in $\Omega$.
Note that $a$ is a critical point of $f$ if $\nabla f(a)=0$, and a critical point $a$ is said to be non-degenerate if $\det H_f(a) \neq0$. Here $H_f$ denotes the Hessian matrix of $f$.
A few observations: $\overline{\Omega}$ and $\partial \Omega$ are both compact. Hence $f$ obtains its extreme value in $\overline{\Omega}$, and the restriction of $f$ i.e. $f|_{\partial\Omega}$ also obtains its extreme value. And, take maximum value for example, $\max_{x\in \partial\Omega} f(x) \leq \max_{x\in \overline{\Omega}} f(x)$. Need help.

Answer 1

Create a cover of your set $\bar \Omega$: The set $R=\{x\in \bar \Omega : \nabla f(x)\neq 0\}$ is open (because the function is $C^1$), then the function $\det H_f(x)$ has a sign (because $f\in C^2(\Omega) )$ in some small ball $B_{\epsilon_i}(x_i)$ centred in the nondegenerate critical point $x_i$. The family $R,\cup_i B_{\epsilon_i}(x_i)$ covers $\bar \Omega$. Now use the compactness to conclude.