Every KC, i.e. "Kompacts are Closed", (and thus every $T_2$) space has the property I'll call IKK: the Intersection of any family of Kompact subsets is itself Kompact.
Not all IKK spaces are $T_1$: consider the right-ray topology on $\omega$, where the non-empty open sets are exactly $\{n,n+1,\dots\}$. In such a $T_0$-not-$T_1$ space, every subset is compact (covering the least element covers the subset), so the space is IKK.
But are all $T_1$ IKK spaces $KC$, or some weaker strengthening of $T_1$? See e.g. https://math.stackexchange.com/a/4761212/ for a few examples of properties between $T_1$ and $T_2$.
Just summarizing from the comments. The finite complement topology on an infinite set is $T_1$, and $IKK$ since indeed, every subset is compact. However, for the same reason, it is not $KC$, since the only nontrivial closed sets are finite.
Moreover, as also remarked in the comments, the finite complement topology on an infinite set fails to satisfy even the weakest property at the linked post, namely the property $US$ (every convergent sequence has exactly one limit), since every sequence that has no duplicate points must converge to every point in the space.
Remark
Conversely, the property $US$ does not imply $IKK$. For a counter-example, consider $[0,\omega_1]$ with a doubled endpoint $\omega_1'$, then $[0,\omega_1]$ and $[0,\omega_1)\cup \{\omega_1'\}$ are compact, but their intersection $[0,\omega_1)$ is not.