Let $(X,\tau)$ be a topological space. A retraction is a continuous map $r:X\to X$ such that $r$ is the identity on $\text{im}(r)$. We call $S\subseteq X$ a retract of $X$ if there is a retraction $r:X\to X$ such that $\text{im}(r) = S$.
We say that $(X,\tau)$ is rc if all retracts are closed. It turns out that all Hausdorff spaces have this property.
Is the class of rc-spaces closed under topological products?
Take $X$ to be an RC space which isn't $T_2$ such as the one-point compactification of the rationals. We will show $X^2$ is not RC. Note that it is not $T_2$ as its factors are not $T_2$.
First we will note that the diagonal $\{\langle x,x\rangle:x\in X\}\subseteq X^2$ is a retract: send $\langle x,y\rangle \mapsto_f \langle x,x\rangle$. This is the composition of the (continuous) projection map onto the first coordinate and the (homeomorphic) map $x\mapsto \langle x,x\rangle$, and thus continuous. Since this map restricted to the diagonal is the identity, it is a retraction.
As noted in this answer, a non-$T_2$ space has a non-closed diagonal: choose $x\not=y$ for which no open neighborhoods are disjoint. Then any basic neighborhood $U\times V$ of $\langle x,y\rangle$ must have $z\in U\cap V$ and therefore $\langle z,z\rangle\in U\times V$, showing $\langle x,y\rangle$ is a limit point of the diagonal.
Thus the diagonal is a retract of $X^2$ which is not closed, showing $X^2$ is not $RC$.
