US does not imply AB

Question

We say that a topological space $X$ is:

1. $AB$, provided that $X$ is $T_1$ and for each pair $(A, B)$ of compact, disjoint subsets of $X$ there is $U$, an open subset of $X$, such that either $A \subseteq U$ and $U \cap B = \varnothing$ or $B \subseteq U$ and $U \cap A = \varnothing$.
2. $US$, provided that convergent sequences have a unique limit.

I would like to know an example of a topological space that is $US$ but not $AB$.

Thanks

Answer 1

Take [0,1] with the usual topology and attach a new point y so that the neighborhoods of y are precisely those which are and open dense in [0,1]. To obtain a 2nd disjoint copy, repeat the construction with [2,3] and a 2nd special point z whose neighborhoods are precisely those which are open and dense in [2,3]. Let A equal the union of [0,1] and z, and let B equal the union of [2,3] and y.

Singletons are closed. There exist no interesting sequences converging to y or z and thus we have a US space. Each neighborhood of A intersects B and vice versa, so this is not an AB space.