While I was reading Do Carmo's differential geometry, I had several questions about the definition of acregular surface.
From condition 2, the author said : "... $x^{-1}:V \cap S \rightarrow U$ which is continuous; that is, $x^{-1}$ is the restriction of a continuous map $F:W \subset \mathbb{R}^3 \rightarrow \mathbb{R}^2 $ defined on an open set $W$ containing $V \cap S$." I know that if $x^{-1}$ is the restriction of a continuous map $F:W \subset \mathbb{R}^3 \rightarrow \mathbb{R}^2$, then $x^{-1}$ is continuous w.r.t. the subspace topology of $V \cap S$. "
However, how can we prove the converse? I.e. if we already know $x^{-1}$ is continuous, how can we show that it is an restriction of a continuous function $F$ which is defined on an open set $W \subset \mathbb{R}^3$?
My second question is concerned with the definition of "differentiable" of condition 1. From MathWorld, it requires $x$ is differentiable – does that 'differentiable' mean $x \in C^{\infty}$?
I'm not familiar with the book, but based on this excerpt, it seems advisable to find a better book on differential geometry. It should not be hard to find something with a more modern perspective on the subject. The repeated references to the ambient space $\mathbb R^3$ are not really necessary.
Concerning your question, I think $W=V$ should work. Indeed, $x^{-1}$ is continuous on the set $V\cap S$, which is closed relative to $V$. (A subset of $V$ is closed in the induced topology if and only if it is the intersection of $V$ with a closed subset of $\mathbb R^3$.) In reasonable spaces such as $V$, every continuous function can be extended from a closed set to the entire space: for a precise statement, see the Tietze extension theorem.