Continuously differentiable map from $\mathbb{R}^{m+n}$ to $\mathbb{R}^n$

Your reasoning using invariant of domain is correct, but overkill for the question you are asking. After all, you did not use differentiability of $f$ at all.

A different approach would be the following: Let $k \leq n$ be the maximal rank of all differentials $Df(x)$ for $x \in U$. It is then not hard to show that the set $$V := \{x\in U\mid \text{rank}(Df(x)) = n\} = \{x\in U\mid \text{rank}(Df(x)) \geq n\}$$

is open (use that for $x\in V$, there is a $k\times k$ submatrix of $Df(x)$ which is invertible).

By assumption, $f$ is injective on $U$ and hence on $V$. But the constant rank theorem (see e.g. Constant rank theorem) shows that $f$ is (on $V$, up to diffeomorphisms) locally of the form $(x_1,\dots ,x_{n+m})\mapsto (x_1,\dots , x_k)$, which is not injective.