Let U be an open subset of $\mathbb{R}^2$ and let K be a compact subset of U. Suppose that f : U → R is a function of class $\mathbb{C}^1$(U) -continuously differentiable in U- and let $\Bbb E = \{(x, y)\in K \mid f(x, y) = 0 \}$ and that Df -jacobian of f- does not vanish on E. Investigate whether E is a Jordan region.
Since determinant of the jacobian matrix of f is not equal to zero in E, initially it came to my mind to use the inverse function theorem. Let's define F(x,y)=( f(x,y), 1) for any $(x,y) \in U$. F is continuously differentiable. Since for any $(x,y) \in E$ the jacobian of f is different than zero, then (by inverse function theorem) there is an open set W containing E such that F is injective on W and $F^{-1}$ is continuously differentiable on F(W). Thus E$ \subseteq$W, E=$F^{-1}((0,1))$ and (0,1) is a point of volume zero. Due to the fact that $F^{-1}$ is continuously differentiable on a compact set K, $E\subseteq K$, then for any region of volume zero its image under $F^{-1}$ will be zero, i.e., E=$F^{-1}((0,1))$ is of volume zero.
Is this proof sufficient? I have some doubts especially about showing that image of (0,1) under F$^{-1}$ is equal to E.