This is not a complete answer, just a possibly useful observation.

**Theorem:** If the property $T_\mathrm{rc}$ is preserved by finite products then any $T_\mathrm{rc}$ space is $T_2$.

*Proof.*
Suppose $T_\mathrm{rc}$ is preserved by finite products and $X$ is $T_\mathrm{rc}$. By assumption $X \times X$ is also $T_\mathrm{rc}$, therefore its diagonal $\Delta_X = \{(x,y) \in X \times X \mid x = y\}$ is closed, as it is a retract of $X \times X$ by the map $(x,y) \mapsto (x,x)$. But a space is $T_2$ iff its diagonal is closed, so $X$ is $T_2$.