Transversal sections of open and closed sets in the product topology. Problem: Let be $X$ and $Y$ topological spaces and $O\subseteq X\times Y$. 
$(i)$ Suppose that $O$ is an open set, where $X\times Y$ have the product topology. Then $p_X\left( X\times\left\lbrace y\right\rbrace\cap O \right)$ is open in $X$, where $p_X:X\times Y\ni (x,y)\mapsto x\in X$ is the projection in the first component. 
$(ii)$ Supose that $O$ is closed. Then $p_Y\left(\left\lbrace x\right\rbrace\times Y\cap O \right)$ is closed in $Y$, where $p_Y:X\times Y\ni (x,y)\mapsto y\in Y$ is the projection in the second component. 
Is known that, the projection functions are open and, in general aren´t closed, but this fact It does not seem to help much here, because  $X\times\left\lbrace y\right\rbrace\cap O$ is not necessarily open. 
I do not find counterexamples of these statements either.  
 A: Let $x \in p_X[(X \times \{y\}) \cap O]$, which is true iff $(x,y) \in O$. Then let $U \times V$ (where $U \subseteq X, V \subseteq Y$ are open) such that $(x,y) \in U \times V \subseteq O$. Note that this implies that for any $p \in U$, we have that $(p,y) \in U \times V$, hence $(p,y) \in O$ so $p \in p_X[(O \times \{y\}) \cap O]$. So $U$ witnesses that $x$ is an interior point of $p_X[(X \times \{y\}) \cap O]$. As $x$ was arbitrary, $p_X[(X \times \{y\}) \cap O]$ is open.
Suppose $C$ is closed (I prefer not to call my closed sets $O$) and let $D = p_Y[(\{x\} \times Y) \cap C]$ and try to show $D$ is closed.
So let $q \in Y$ be such that $q \notin D$. This means that $(x,q) \notin C$.
As $C$ is closed we find a basic open set $U \times V \ni (x,q)$ that is disjoint from $C$. Then $V$ contains $q$ and is disjoint from $D$ (otherwise $q' \in D \cap V$ exists and $(x,q') \in C$, but $(x,q') \in (U \times V) \cap C$ which is empty by construction). So every point not in $D$ has an open neighbourhood that is disjoint from $D$, so $D$ is closed.
