Let $f: R^2\to R^2$ be a $C^1$ map such that $f^{-1}(y)$ is finite for all $y\in R^2$. Prove that $\det df(x)$ cannot vanish on an open set.
My thoughts: Suppose the determinant vanishes on an open set. Then it vanishes locally on every disc in the set. Any disc is convex, so $f$ is constant on that disc by the corollary to Rudin's Theorem 9.19 (Mean Value Theorem). So $f$ is locally constant. Let $D$ be one of the discs, $x\in D$. Then $f^{-1}(f(x))$ contains the entire disc $D$, a contradiction.
Is this argument valid?