topology on a compact hausdorff not comparable? The original problem:
Show that if $X$ is compact hausdorff under both $\tau$ and $\tau'$, then either $\tau = \tau'$ or they are not comparable?
Most answer on the internet base on this logic:
Assuming they are comparable, we can prove they must be equal. Then end their proof. (which I think is not complete)
My question is how to show there are two topology that are not comparable on $X$ and both make $X$ compact hausdorff?
Can anyone give some examples ?
 A: A very simple example is to let $X=\Bbb N$ and define the following topologies. (Note that for me $0\in\Bbb N$.) Let
$$\tau_1=\wp(\Bbb Z^+)\cup\{U\subseteq\Bbb N:0\in U\text{ and }\Bbb N\setminus U\text{ is finite}\}\;;$$
the only non-isolated point is $0$, and every nbhd of it contains all but finitely many points of $\Bbb N$, so $\langle\Bbb N,\tau_1\rangle$ is compact and Hausdorff. Now let $E=\{2n:n\in\Bbb N\}$, the set of even non-negative integers, and let $O=\Bbb N\setminus E$, the set of odd positive integers. Let $\tau_2$ be the topology generated by the following base:
$$\begin{align*}
\big\{\{n\}:n\in\Bbb N\setminus\{0,1\}\big\}&\cup\{U\subseteq E:0\in E\text{ and }E\setminus U\text{ is finite}\}\\
&\cup\{U\subseteq O:1\in E\text{ and }O\setminus U\text{ is finite}\}\;.
\end{align*}$$
Then $\{0\}\cup E$ and $\{1\}\cup O$ are clopen subspaces of $\langle\Bbb N,\tau_2\rangle$ that are homeomorphic to $\langle\Bbb N,\tau_1\rangle$, so $\langle\Bbb N,\tau_2\rangle$ is also compact Hausdorff. However, $\langle\Bbb N,\tau_2\rangle$ has two non-isolated points, while $\langle\Bbb N,\tau_1\rangle$ has only one, so the two spaces are not homeomorphic.

As for the original theorem, assuming that two compact Hausdorff topologies on a set are comparable and proving from that assumption that they must be equal is a complete proof that any two compact Hausdorff topologies on a set are either incomparable or equal; that’s elementary logic.
A: You are somewhat right about the completeness of the proof, but the formal argument is complete. Indeed, there are 3 possibilities: the two topologies are equal (E), or they are different (D), in which case they can be either comparable (DC) or not comparable (DN). You need to prove that if they are both compact and Hausdorff then only E and DN are possible. It is equivalent to show that if they are comparable (C) then they are equal, because by this you rule out DC and show that only E or DN are possible. 
What you are right about is the informal part of this statement, namely the question that arises naturally next: is DN possible at all? So, again, this question is not a part of the argument, but rather a question you would ask next. 
There are many ways to construct such examples. It might seem more difficult to find a good example of a compact topology on a given set, but then you have the compactification theorem which basically tells you that all you need is a locally compact space, and there are plenty of them. In fact, you need to think to find one that is not locally compact. 
So, for example, take a LCH (locally compact Hausdorff) space X. Its compactification we will call X'. Then, two different LCH topologies on X give you two different compact Hausdorff (CH) topologies on X', and here you go. 
Or, take just one LCH X, and take X''. If X is infinite, X' and X'' have the same cardinality, so you can assume they are on the same space Y. Then, it is easy to make their topologies different by choosing, for example, discrete X. 
