EDIT: Is it rigorous to define the open set $A=(a,b)\times (c,d)$? I think to prove $g$ is an open map, we need to pick an arbitrary open set in $\mathbb{R}^2$, but this open set needs not having the form $(a,b)\times (c,d)$. For example, it can be $A=(a,b)\times (c,d)\cup(e,f)$. Am I correct?
Can someone explain the circled part of the proof? To apply Corollary 22.3, don't we need to prove $X^*=\{g^{-1}(z):z\in\mathbb{R}\}$? The author just used one sentence to finish this part? The question said $X^*$ be a corresponding quotient space. Corresponding to a quotient map $p$ or $g$?
$X^*$ is a collection of subsets of $\mathbb{R}\times\mathbb{R}$ such that two pairs $(x_0,y_0),$ $(x_1,y_1)$ belong to the same element of $X^*$ if and only if $$x_0+y_0^2=x_1+y_1^2 \in \mathbb{R},$$ and that happens exactly when $g(x_0,y_0)=g(x_1,y_1) \in \mathbb{R},$ that is, if and only if $(x_0,y_0)\in g^{-1}(z)$ and $(x_1,y_1)\in g^{-1}(z)$ for the same $z\in\mathbb{R}.$ Hence $$X^*=\{g^{-1}(z)\text{ | }z\in \mathbb{R}\}.$$