(Here I am using the terminology of the book, see my comment under the question.)
Assume for every open set $U\subset {\mathbf R^m}$ there is some open set $V\subset {\mathbf R^n}$ such that $f^{-1}(U)=V\cap A$. Let $a\in f^{-1}(U)$, where $U$ is open. Because $U$ is open, there is some open rectangle $B$ with $f(a)\in B\subset U$. Then $f^{-1}(B)$ is open in ${\mathbf R^n}$, thus for any interval of size $\epsilon$ in ${\mathbf R^m}$ we can find an interval of size $\delta$ in $A$ such that if $|x-a|<\epsilon$, $|f(x)-f(a)|<\delta$.
The statement of the proof looks like it is saying more than that, but it isn't. Basically it just says that if the inverse image of an open set is an open set, we can wangle that into the epsilon delta definition of continuity. As the book says that is a "pleasant surprise".