Need a unique convergence (UC) space's Alexandrov extension be a UC space? 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$_1$.
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.
I've been able to show that $X$ is P iff $X_\alpha$ is P, when P is KC, UCC, or T$_1$. I'm also aware that $X_\alpha$ is Hausdorff iff $X$ is Hausdorff and locally compact.

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. Clearly, any counterexample would need to not be a UCC space, but all the UC, non-KC spaces I've seen or been able to come up with are also UCC, so I'm not having any luck on that front.
If it is true, how might one go about proving it?
If it is not true, can you provide a counterexample? As a bonus question, are there conditions we can add so that we can obtain an equivalence, such as the locally compact condition in the Hausdorff case? Local compactness would certainly be sufficient for a UC space to have a UC Alexandrov extension, but is it necessary?
Added: I do know that a counterexample cannot be first-countable, since first-countable UC spaces are UCC. And of course, it can't be compact, or even locally compact.
 A: 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 $q$.
