Prove that a finite set has exactly one Hausdorff topology I have never seen this problem before unfortunately, until I came across in a past paper.

Prove that a finite set has exactly one Hausdorff topology.

My approach was first to actually assume that there are two distinct topologies $\tau_1,\tau_2$ such that $(X,\tau_1),(X,\tau_2)$ are Hausdorff topologies.
Then my aim is to somehow show that elements of each topologies are exactly the same for it to be Hausdorff but I am already stuck here and I don't have any other ideas to tackle this.
Does anyone have a solution to this?
 A: Hint: prove that if $\tau$ is any Hausdorff topology on a finite $X$, then $X = \mathscr{P}(X)$, the discrete topology. This clearly shows uniqueness.
Use that a finite intersection of open sets is open, so here all intersections of open sets are open (as there are at most finitely open sets in any topology on a  finite set).
A: Instead of equating two arbitrary topologies, you might try equating an arbitrary topology with one you already know, namely here the discrete topology (where every subset is open). 
For this, take a topology $\tau$ on the finite set $X$, and let's prove that singletons are open. Take a point $x \in X$. For any $y \ne x$, there exist $O_y \in \tau$ a neighborhood of $x$ that does not contain $y$. Now consider the set 
$$O = \bigcap_{y \ne x} O_y$$
The set contains $x$ (since all the $O_y$ do), is open (as a finite intersection of open sets) and does not contain any of the $y\ne x$. Therefore it is the singleton ${x}$ and we thus  have proved that singletons are open. To conclude, note that any set is an union of singletons and is therefore open.
