Background
Say a topological space $X$ is
- a unique convergence (UC) space iff every sequence of points of $X$ converges to at most one point of $X$;
- a unique convergent clustering (UCC) space iff every convergent sequence of points of $X$ has a unique cluster point;
- a KC space iff compact subsets of $X$ are closed.
I've been able to show that Hausdorff $\implies$ KC $\implies$ UCC $\implies$ UC $\implies$ $T_1,$ and that a first-countable UC space is UCC.
Given a space $X$, we define the Alexandrov extension of a space $X$--which I denote $X_\alpha$--by adjoining a point $\infty$ to $X$, and making the neighborhoods of $\infty$ those subsets of $X\cup\{\infty\}$ whose complements are closed compact subsets of $X$. It can be shown that $X_\alpha$ is always compact, $X$ an open subspace of $X_\alpha,$ and that $X$ is dense in $X_\alpha$ if and only if $X$ is non-compact.
It is trivial to show that $X$ is $T_1$ iff $X_\alpha$ is $T_1,$ since finite sets are immediately compact. I'm also aware that $X_\alpha$ is Hausdorff iff $X$ is Hausdorff and locally compact.
Edit: I also "proved" before posting the question that $X$ is KC iff $X_\alpha$ is KC, but this turns out not to be true. When $X$ is the Arens-Fort space, then $X$ is Hausdorff (so KC), but $X_\alpha$ is not even KC, as pointed out in the comments. To see why, note that every neighborhood of the "point at infinity" in $X_\alpha$ is cofinite, so every subset of $X_\alpha$ containing this point is compact. However, the set consisting of "even columns" of $X$ is not open in $X_\alpha$ and doesn't contain the "point at infinity," so its complement is a compact, non-closed subset of $X_\alpha.$ Thanks again to MW for bringing that lovely counterexample to my attention!
Edit: Another "proof" was that $X$ was UCC if and only if $X_\alpha$ was UCC, but again, the Arens-Fort space provides a counterexample (so is an even stronger counterexample to preservation of the KC property by Alexandrov extension), as pointed out in the comments once again by MW.
It is readily the case that if $X_\alpha$ is a UC space, then so is $X$. However, I've been unable to prove or disprove that if $X$ is UC, then so is $X_\alpha$. I suspect that it isn't true in general, but I haven't found any counterexamples.
If it is true, how might one go about proving it?
If it is not true, can you provide a counterexample?
Here’s a counterexample.
Let $P=\beta\omega\setminus\omega$ be the set of free ultrafilters on $\omega$. Let $X=(\omega+1)\cup P$ be topologized by making $\omega+1$ an open subset of $X$ with its usual order topology and declaring $U\subseteq X$ a nbhd of $p\in P$ iff $p\in U$ and $U\cap\omega\in p$. The only non-trivial sequences in $X$ are those that converge to $\omega$, so $X$ is $US$, but $\langle n:n\in\omega\rangle$ clusters at each point of $P$ (and of course converges to $\omega$), so $X$ is not $UCC$. Note that $\omega$ is dense in $X$, which is not compact, so it’s not the case that the range of a convergent sequence in a $US$ space must have compact closure.
$P$ is an infinite closed, discrete subset of $K$, and $P\cap\operatorname{cl}A$ is infinite for each infinite $A\subseteq\omega$. (In fact $|P\cap\operatorname{cl}A|=|\beta\omega\setminus\omega|=2^{\mathfrak{c}}$ for each infinite $A\subseteq\omega$.) Thus, the only compact, closed subsets of $X$ are the finite sets, and if $q$ is the point at infinity in $X^*$, the sequence $\langle n:n\in\omega\rangle$ converges to both $\omega$ and $q$.