A function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is said to be proper if it is continuous and, for every compact $K \subset \mathbb{R}^m$, $f^{-1}(K)$ is a compact set in $\mathbb{R}^n$.
$1)$ Show that every function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, continuous, injective and closed is proper.
$2)$ Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ a $\mathcal{C}^1$ function such that the jacobian matrix of $f$ at each point of $\mathbb{R}^n$ is invertible. Show that $f$ is proper.
So, I've solved the number $1$ and I don't know how to solve number $2$. My attempt for number $2$ was, since the jacobian matrix is invertible, its determinant is $\neq 0$. Now, the function being $\mathcal{C}^1$, from the inverse function theorem we can find "locally" $\mathcal{C}^1$ diffeomorphisms and each of these diffeomorphisms are proper, from number $1$. And this was everything I've done for number $2$, I was trying to use the ideia of open cover to finish, but didn't succeed. So, how to finish this question? Or, how to do this question?
For number $1)$
Let $K \subset \mathbb{R}^m$ a compact set, since $f$ is continuous, $f^{-1}(K)$ is closed, because $K$ is closed. Now we have to show that $A = f^{-1}(K)$ is bounded. Suppose by contradiction that $A$ is not bounded, so we can find a sequence $\{x_n\}_{n \in \mathbb{N}} \subset A$ such that $\|x_n\| \rightarrow \infty$ as $n$ goes to $\infty$. The function is closed, so $f(A)$ is closed, but $f(A) \subset K$, so $f(A)$ is compact. $\{f(x_n)\} \subset f(A)$ and $f(A)$ compact we can find a convergent subsequence, wlog consider this subsequence equal to $\{f(x_n)\}$, so $f(x_n) \rightarrow \overline{x}$, for some $\overline{x}$, but $f(A)$ is closed, so $\overline{x} \in f(A)$, so there is a $x \in A$ such that $f(x) = \overline{x}$, this $x$ is unique because $f$ is injective. Now consider the "restriction" $f: \mathbb{R}^n \rightarrow f(\mathbb{R}^n)$, this function is a bijection because $f$ is injective and it is cleary surjective. Its inverse $f^{-1}$ is continuous because the inverse image of closed set is closed, if $F \subset \mathbb{R}^n$ is a closed set, then $(f^{-1})^{-1}(F) = f(F)$ is closed because $f$ is closed. Now, by continuity, $f(x_n) \rightarrow f(x) \Rightarrow x_n = f^{-1}(f(x_n)) \rightarrow f^{-1}(f(x)) = x$, a contradiction. So $A = f^{-1}(K)$ is bounded, and, since closed and bounded is equivalent to compact in $\mathbb{R}^n$, $f^{-1}(K)$ is compact. Because of it we have that $f$ is proper.
For number $2)$
This question is false, just consider the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, $f(x, y) = (e^x \cos y, e^x \sin y)$. $f$ is $\mathcal{C}^1$ and has invertible jacobian matrix in every point of $\mathbb{R}^n$, but $f$ is not proper, we can see that $f^{-1}(\{(1, 0)\})$ is not bounded, so it can't be compact.