"All retracts are closed" and "all compacts are closed" I want to follow the discussion from here 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.
 A: RC does not imply KC: in this paper 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 $T_2$.
A: EDIT: This answer relied on an accepted answer elsewhere that has now been updated to remove an oversight. See my note below.
First I need to prove that the Arens-Fort space $X$ is not compactly generated. To do this, I'll show there exists a non-closed set which has closed intersection with every compact. This is very easy: since this space is not discrete, it has an infinite non-closed set $C$. Since this space is anticompact, all compacts are finite, and thus $C$ intersected with every compact is finite. Finally, since the space is Hausdorff, $C$ intersected with every compact is closed.
Theorem 5 of Between $T_1$ and $T_2$ shows us that $X$'s one-point compactification $X^+$ is not $KC$ since $X$ is not compactly generated.
Finally, $X^+$ is $RC$, since it fits the requirements of Paul Fabel's answer here.
Thus $RC$ does not imply $KC$, even for compact spaces.

EDIT: $X^+$ is not $RC$; Fabel's answer has now been updated to require $X$ to be compactly generated, which our $X$ is not.
To see this directly, let $0$ be the non-isolated point of $X$; let $\infty$ be the new point in $X^+$.
Then $A=X^+\setminus\{0\}$ is non-closed; we claim it is a retract. Let $f:X^+\to A$ be defined by $f(0)=\infty$ and $f(x)=x$ otherwise. Note $f\upharpoonright A=id_A$ and $A$ is open, so $f$ is continuous at each point of $A$. Let $U$ be a neighborhood of $f(0)=\infty$. Since $X$ is anticompact, $U$ is cofinite. Thus $f^\leftarrow[U]$ is also cofinite, and thus open, proving $f$ is continuous at $0$. This completes the proof of our claim.
And just to round this out, $A$ also provides an explicit example of why $X^+$ is not $KC$: it's compact as every neighborhood of $\infty$ is co-finite.
