Do Carmo differential geometry book: section 2.2 proposition 4 proof clarification I am confused about this proposition:

Prop: Let $p \in S$ be a point of a regular surface $S$, and let ${\bf x}: U \subset R^2 \rightarrow R^3$ be a map with $p \in x(U)$ such that the conditions 1 and 3 of the definition 1 hold. Assume $x$ is one to one. Then ${\bf x}^{-1}$ is continuous.

Definition 1 is:

A subset $S \subset R^2$ is a regular surface if, for each $p \in S$, there exists a neighborhood $V$ in $R^3$ and a map ${\bf x}: U \rightarrow V \cap S$ of an open set $U$ in $R^2$ such that:
1) ${\bf x}$ is differentiable.
2) ${\bf x}$ is a homeomorphism.
3) For each $q \in U$, the differential $d{\bf x}_q : R^2 \rightarrow R^3$ is one to one.

Essentially they claim that the continuity of the inverse map is automatic once other conditions are satisfied once we know that $S$ is already a surface. But I do not see how they make use of the fact that $S$ is a surface anywhere in the proof. Also in their proof, they show that each point in the image of ${\bf x}$ has a set around it (no claim on it being open) on which the inverse function is continuous, and conclude that it is enough.  Is it?
EDIT: (adding the actual proof they present)

Write ${\bf x}(u, v) = (x(u, v), y(u, v), z(u, v))$, for $(u, v) \in U$, and let $q \in U$.
  By conditions 1 and 3, we can assume, interchanging the coordinate axis
  if necessary, that $(\partial(x, y)/\partial(u, v))(q) \neq 0$. Let $\pi: R^3 → R^2$ be the projection
  $\pi(x, y, z) = (x, y)$. By the inverse function theorem, we obtain neighborhoods $V_1$ of $q$ in $U$ and $V_2$ of $\pi ◦ {\bf x}(q)$ in $R^2$
  such that $\pi ◦ {\bf x}$ maps $V_1$
  diffeomorphically onto $V_2$.
Assume now that ${\bf x}$ is one-to-one. Then, restricted to ${\bf x}(V_1)$, ${\bf x}^{−1} = (\pi ◦ {\bf x})^{−1} \circ \pi$ . Thus ${\bf x}^{−1}$, as a composition of continuous maps, is continuous. Since $q$ is arbitrary, ${\bf x}^{-1}$ is continuous in ${\bf x}(U)$.

 A: You are a clever student. While the statement of the proposition is true, in the versions of Do Carmo that I have (the most recent is from 2016), the proof failed to establish the truth. The method of the proof works perfectly for an injective, regular parametrization of the figure-eight curve in the plane and does indeed show that the parametrization defines homeomorphisms from smaller open intervals to pieces of the curve. The reason you cannot deduce the global inverse is continuous (and the way it is different from what happens when the image is inside a regular surface) is that these pieces of the figure-eight curve are not open sets in the topology of the figure-eight curve. This is the part of the proof that is lacking: Do Carmo never shows that the parametrization in question is an open mapping (in terms of the surface). Although this fact can be established under the hypothesis of Proposition 4, it is not necessarily true for an injective, regular parametrized surface (e.g. a cylinder over the figure-eight curve) in terms of the global image. Don't get me wrong, I think Do Carmo has written an outstanding book, but considering how much is has been used, I'm surprised how long this flaw has survived.
