I use the following definition of $T_4$-space: for any two disjoint closed sets $A$, $B$ there exist disjoint open sets $U$, $V$ containing $A$ and $B$ respectively.
Is it true that factor spaces are $T_4$ if product space is $T_4$?
This is obviously true for normal spaces ($T_1$ and $T_4$), but i can't find any proof for "$T_4$ only" case. If this is not true, I need a counterexample.
It suffices to consider the case of a product of two factors, so let $X,Y$ be nonempty topological spaces, and suppose $X\times Y$ is a $T_4$ space.
Let $A,B \subset X$ be disjoint closed sets. Then $A\times Y$ and $B \times Y$ are disjoint closed sets in $X \times Y$, so there are disjoint open $U, V \subset X\times Y$ with $A\times Y \subset U$ and $B\times Y \subset V$. Choose an $y_0 \in Y$, and consider the injection $\iota \colon x \mapsto (x,y_0)$. Since $\iota$ is continuous, $\iota^{-1}(U)$ and $\iota^{-1}(V)$ are open. They are disjoint since $U \cap V = \varnothing$, and we have $A \subset \iota^{-1}(U)$, $B \subset \iota^{-1}(V)$. Thus $X$ is a $T_4$ space.
The same argument works for $Y$ of course.