A finite Hausdorff space is discrete 
Theorem: $X$ is a finite Hausdorff. Show that the topology is discrete.

My attempt: $X$ is Hausdorff then $T_2 \implies T_1$ Thus for any $x \in X$ we have $\{x\}$ is closed. Thus $X \setminus \{x\}$ is open. Now for any $y\in X \setminus \{x\}$ and $x$ using Hausdorff property, we get $\{x\}$ is open. 
Am I right till here? And how to proceed further? 

You're a bit sloppy in assuming that $\{x\}$ is open.
The thing you have to prove is that any subset of $X$ is open. This is quite straight forward as every subset of $X$ is $X$ minus a finite number of points, if it's not $X$ itself (which is open anyway) it's minus a finite positive number of points. That is you can write the subset as a finite intersection:
$$\bigcap X\setminus \{x_j\}$$
but the set $X\setminus \{x_j\}$ is open as you pointed out. And it's known that finite intersection of open sets is open. So any subset of $X$ is therefore open.
The same reasoning can be used to specially prove that $\{x\}$ is open, but we can prove the topology to be discrete directly here.
