Let $q:X\rightarrow Y$ be an open quotient map and $A\subseteq X$ be saturated. Then $intA$ is saturated.
My attempt:Let $A$ be a saturated subset of $X$. So $A=q^{-1}(q(A))$. I claim that $int(q^{-1}(q(A))$ $=$ $q^{-1}(int(q(A))$ (This, if it holds, then by definition implies that intA is saturated).
If $x\in int(q^{-1}(q(A))$ then we can find a neighborhood U of $x$ such that $U\subseteq q^{-1}(q(A))$. Which implies that $q(U) \subseteq q(A)$. Since $q$ is an open map, $q(U)$ is open and contains $q(x)$. Thus, $q(x)\in int(q(A))$ and so $x\in q^{-1}(int(q(A))$. For the reverse inclusion, if $x\in q^{-1}(int(q(A)))$ then $q(x)\in int(q(A))$. This means that we can find a neighborhood $V$ of $q(x)$ such that $V\subseteq q(A)$. Hence, $q^{-1}(V)\subseteq q^{-1}(q(A))$. Since $q^{-1}(V)$ contains $x$, and is open by continuity of quotient maps, it follows that $x\in int(q^{-1}(q(A)))$. Since $x$ was arbitrary, $int(q^{-1}(q(A))$ $=$ $q^{-1}(int(q(A))=int(q(A))$.
Note the equivalence:
Definition: If $f:X\rightarrow Y$ is a map and $A\subseteq X$. Then $A$ is called saturated if $A=q^{-1}(V)$ for some $V\subseteq Y$.
If $f:X\rightarrow Y$ is a map, $A\subseteq X$ is saturated if and only if $A=q^{-1}(q(A))$.
Is my proof correct? (Please answer this first).
Alternative proof idea: If $A$ is saturated, $A=q^{-1}[B]$ for some $B \subseteq Y$.
Try to show that $\operatorname{int}(A)= q^{-1}[\operatorname{int}(B)]$ using openness and continuity of $q$. Use that $\operatorname{int}(A)$ is by definition the largest open subset of $A$. Then it's directly seen to be saturated.