Assume $\Omega \subset \mathbb{R}^m$ is a connected bounded open set, function $f\in C^1(\Omega,\mathbb{R}^m$) satisfies
(1) $det f'(x) \neq 0,\forall x\in \Omega$
(2) $\forall x_0\in \partial \Omega$ $lim_{x\to x_0}\vert f(x)\vert =\infty$
We can prove that for all $y\in f(\Omega)$ ,$f(x)=y$ has only finite different roots by inverse function theorem and open covering theorem. And there is a stronger theorem: denote $f^{-1}y=\{x\in \Omega \vert f(x)=y\}$,$\Phi(y)=\vert f^{-1}(y)\vert<\infty$,
How to prove $\Phi(y)$ is a constant function?
The stronger theorem can be deduced from the weaker theorem that you can already prove, by applying the definition of connectivity, together with a further application of the inverse function theorem, and some more work.
The way that connectivity will be applied is to prove that the function $\Phi : f(\Omega) \to \mathbb N = \{1,2,3,...\}$ is continuous. But $\Omega$ is connected, so $f(\Omega)$ is connected, so $\Phi(f(\Omega))$ is connected, and every nonempty, connected subset of $\mathbb N$ is a single point, hence $\Phi$ is constant.
Consider $y \in f(\Omega)$. Our goal is to find an open neighborhood of $y$ on which the function $\Phi$ restricts to a constant.
Applying the weaker theorem, the set $f^{-1}(y)$ is finite. Let $K=\Phi(y)$, and let's enumerate $$f^{-1}(y) = \{x_1,...,x_K\} $$ For each $k=1,...,K$, by applying the inverse function theorem one obtains open neighborhoods $U_k \subset \Omega$ of $x_k$, and $V_k \subset f(\Omega)$ of $y \in V_k$, such that $f$ restricts to a $C^1$-diffeomorphism $f : U_k \to V_k$; that restriction is therefore a bijection.
Define $V = V_1 \cap \ldots \cap V_K$, and so $V \subset f(\Omega)$ is an open neighborhood of $y$. For each $k=1,...,K$ let $W_{k} = U_k \cap f^{-1}(V)$, and so $W_k$ is a neighborhood of $x_k$ and $f$ restricts to a $C^1$-diffeomorphism $f \mid W_{k} \to V$.
Letting $W = W_1 \cup ... \cup W_K$, we know so far that the restriciton $f : W \to V$ is exactly $K$-to-$1$. If we knew that $W = f^{-1}(V)$ then we would know that $\Phi \mid V$ takes on the constant value $K$. But unfortunately the equation $W = f^{-1}(V)$ does not have to be true, so we have to work harder.
Choose $R>0$ so that $B(y,R) \subset V_y$. For each $r \ge R$ let $W_{r,k} = f^{-1}(B(y,r)) \cap W_{k}$, and so $f$ restricts to a diffeomorphism $f : W_{r,k} \to B(y,r)$. Notice that for each $r \in (0,R)$ we have an inclusion $$W_{r,1} \cup ... \cup W_{r,K} \subset f^{-1}(B(y,r)) $$ Now let's prove that there exists $r \in (0,R)$ such that $$f^{-1}(B(y,r)) \subset W_{R,1} \cup ... \cup W_{R,K} $$ If we can find such an $r$ then we will be done for the following reasons: the reverse inclusion $$f^{-1}(B(y,r)) \subset W_{r,1} \cup ... \cup W_{r,K} $$ will be true, and therefore the equation $$W_{r,1} \cup ... \cup W_{r,K} = f^{-1}(B(y,r)) $$ will be true, and it will follow that $\Phi$ has constant value $K$ on the open ball neighborhood $B(y,r)$ of $y$.
If $r$ does not exist as described then there would exist a sequence $p_i \in f^{-1}(B(y,R))$ such that $f(p_i) \to y$ as $i \to \infty$ and yet $p_i \not\in W_{R,1} \cup ... \cup W_{R,K}$. Since $p_i$ is bounded, it would have a convergent subsequence, and replacing it by that subsequence we could assume that $p_i$ converges to some $q \in \mathbb R^m$.
If $q \in \Omega$ then $f(q)=f(\lim p_i)=\lim f(p_i)=y$ and so $q = x_k$ for some $k=1,...,K$. The sequence $p_i$ is therefore eventually contained in the neighborhood $W_{R,x}$ of $q=x_k$, a contradiction.
If $q \not\in \Omega$ then $q \in \partial \Omega$. But then $\lim |f(p_i)| = |y| \ne \infty$, also a contradiction.