TITLE: Uniqueness of limits and compactness implies closure
QUESTION [6 upvotes]: It is not difficult to prove that in a Hausdorff topological space every compact set is closed, and almost trivial that if in a topological space X every compact set is closed then X is T1. As uniqueness of limits for a topological space lies between the T1 and T2 properties, it is a candidate for a necessary and sufficient condition for every compact set to be closed. Is it necessary or sufficient?

REPLY [3 votes]: A KC space is indeed US. This is classical. I believe its already in the paper where these notions (KC = all compact subsets are closed, US = all convergent sequences have unique limits) are defined (this paper by Wilansky, I believe).
A proof can be found here, but I'll recap the idea in case the link
goes away: suppose $x_n \to x$ and $x_n \to y$, we need to show that $x=y$.
Suppose $x \neq y$. One of the sets $N(x):=\{n : x_n = x\}$ or $N(y):= \{n: x_n = y\}$ is not cofinite, say the second is not. Omit all terms $x_n$ with $x_n =y$ and so create   a new sequence $y_n$ that still converges to $x$ and $y$. The set $C:=\{y_n\} \cup \{x\}$ is compact (hence closed as we assume $X$ is KC)  but $y$ is also in the closure of $C$ but not in $C$, contradiction and thus $x=y$, and $X$ is US.
This old question and its answers give ideas to find a US space that is not KC.