In Step 2 of Theorem 22.1 from Munkres' Topology:
Let $p:X\to Y$ be a quotient map; let $A$ be a subspace of $X$ that is saturated with respect to $p$; let $q:A\to p(A)$ be the map obtained by restricting $p$.
(1) If $A$ is either open or closed in $X$, then $q$ is a quotient map.
(2) If $p$ is either an open map or a closed map, then $q$ is a quotient map.
the proof verifies that $q^{-1}(V)=p^{-1}(V)$ for $V\subset p(A)$ and $p(U\cap A) = p(U)\cap p(A)$ for $U\subset X$, and then states:
Now, suppose $A$ is open or $p$ is open. Given the subset $V$ of $p(A)$, we assume that $q^ {-1}(V)$ is open in $A$ and show that $V$ is open in $p(A)$.
I don't find the converse proof: We assume that $V$ is open in $p(A)$ and show that $q^ {-1}(V)$ is open in $A$ (Why Munkres does not give this proof?).
We try the following: Let $V$ open in $p(A)$. Since $p$ is a quotient map, $V = p(U) \cap p(A)$ for some set $U$ open in $X$. Then, using the equation from Step 1, $q^{-1}(V) = p^{-1}(V) = p^{-1}(p(U) \cap p(A)) = p^{-1}(p(U \cap A)) = U \cap A$, which is open in $A$.
Is this proof ok?