**This is a community wiki of the answers in the comments.**

* The compact Hausdorff topologies do not generally form a maximal antichain. If X is infinite, split X into two infinite halves and put the discrete topology on one half and the indiscrete topology on the other half. (Comment by François G. Dorais) **Addendum:** Without sufficient Choice, the infinite set $X$ may be [amorphous](http://en.m.wikipedia.org/wiki/Amorphous_set). Amorphous sets are *precisely* the infinite sets for which this approach doesn't work. Very little Choice is needed to ensure that no such beast exists. (Edit by Cameron Buie)

* There is a maximal compact topology on a countable space which is not  Hausdorff. See <a href="http://books.google.com/books?id=DkEuGkOtSrUC&lpg=PA118&ots=3hHBTJE-k7&dq=maximal%20compact%20topology&pg=PA118#v=onepage&q=&f=false">Steen & Seebach 99</a>. (Comment by Gerald Edgar)

* There is a minimal Hausdorff topology on a countable space which is not compact. See <a href="http://books.google.com/books?id=DkEuGkOtSrUC&lpg=PA118&ots=3hHBTJE-k7&dq=maximal%20compact%20topology&pg=PA119#v=onepage&q=&f=false">Steen & Seebach 100</a>. (Comment by François G. Dorais)

* Those examples can be lifted to any cardinality space, simply by using the disjoint sum with any given compact Hausdorff space. (Comment by Gerald Edgar)

* Every set admits a compact Hausdorff topology, by topologizing it as the one-point compactification of the discrete space structure on the complement of any point. (Answer below by Cameron Buie)

(*Feel free to edit and expand*)