I want to follow the discussion from [here](https://mathoverflow.net/questions/191016/all-retracts-are-closed-as-separation-axiom) concerning about the strength of the separation "all retract subspaces are closed".


(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|\_A = \mathrm{id}\_A$.)


Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T\_2\Rightarrow \mathrm{KC}$, $T\_2\Rightarrow \mathrm{RC}$ and neither of the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.


So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.



"All retracts are closed" and "all compacts are closed"
RC does not imply KC: in [this paper](https://arxiv.org/abs/2211.12579) Banakh and Stelmakh construct a semi-Hausdorff countable Brown space $X$ which is strongly rigid (and hence $X$ has $RC$) and contains a non-closed compact subset (so, $X$ fails to have $KC$).


This example also shows that $KC$ does not follow from the semi-Hausdorff property, which is intermediate between $T\_1$ and $T\_2$.


