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.

Let $X$ be a finite Hausdorff space. Let $x\in X$. For each $y\not =x\in X$ let $U_y$ and $V_y$ be disjoint open sets with $x\in U_x$ and $y \in V_y$. Set $V=\cup_{y\not = x} V_y$. Then $V$ is open... So $X\setminus V=\{x\}$ is closed.

Thus every point in $X$ is closed. Since $X$ is finite, every point is also open (complement of finite union of closed sets).

We could actually say that every finite $T_1$ space is discrete, since points being closed is equivalent to being $T_1$.