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?

1$\begingroup$ When you say "uniqueness of limits", do you mean the statement that if a sequence converges, then it converges uniquely? Alternatively, please explain the precise statement you are referring to. $\endgroup$ – user44191 Dec 13 '18 at 17:10

$\begingroup$ Yes, when I say uniqueness of limits I mean the property of a topological space that every convergent sequence converges uniquely. $\endgroup$ – Daniel Elessar Dec 13 '18 at 17:11

3$\begingroup$ It's not going to be sufficient, because sequence convergence isn't enough to detect "limits". You should be able to get a counterexample from a "line with two origins" construction on the uncountable ordinal $\omega_1 + 1$. $\endgroup$ – Nate Eldredge Dec 13 '18 at 18:01

2$\begingroup$ I think it's necessary, though, because a convergent sequence together with one of its limits is a compact set, and it is closed only if it contains all the limits of the sequence. $\endgroup$ – Nate Eldredge Dec 13 '18 at 18:03

3$\begingroup$ @user44191: That suggests looking at a Tychonoff plank with two corners; apparently the corner is a limit point, but not the limit of any ordinalindexed sequence. On the other hand, if you say "ultrafilter limits are unique" or "net limits" then this is equivalent to being Hausdorff. $\endgroup$ – Nate Eldredge Dec 13 '18 at 18:19
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.