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? 
 A: 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.
A: 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 $T_1$.
A: Yes, you're completely right (although, you might want to write out your argument for $\{x\}$ being open in a little more detail.) You've shown that every point is open, so it follows form the axioms for a topology that every set, as a union of points, is open. This is precisely the definition of the discrete topology.
Finiteness is used to conclude that every point is open. (It is certainly not true that every infinite Hausdorff space is discrete. Think of $\mathbb R$!) If you write out the argument more carefully, you'll see where finiteness is used.
A: If you showed that, for Hausdorff spaces , all one point sets $\{x\}$ are closed, it also follows that all finite subsets $F = \{x_1,\dots, x_n \}$ are closed, since
$$
F = \bigcup_{j=1}^n \{x_j\} 
$$
will be closed as a finite union of closed sets.
Let's apply this to the situation where $X$ is finite and Hausdorff. For any subset $U \subseteq X$, the complement $X \setminus U$ will be finite and thus closed by the argument above. Hence $U$ is open. 
So all subsets of $X$ are open, which means the topology is discrete.
A: Let $(X,\tau)$ be a finite topological space.  Let $x\in X$.  If the singleton $\{x\}$ is not an open set, then $(X,\tau)$ cannot be Hausdorff.  This is shown as follows. Let $B$ be the intersection of all open subsets of $X$ that include $x$.  Since $X$ is finite, there are a finite number of such subsets, therefore $B$ is open ($B\in \tau$). If $\{x\}$ is not open, $B$ (since it is open) must include, in addition to $x$, another element $y\neq x$. So any open set $U_x$ containing $x$ also includes $y$. So there is no open set $U_y$ containing $y$ that is disjoint from any open set $U_x$ including $x$, even though $x$ and $y$ are distinct elements. Hence $(X,\tau)$ cannot be Hausdorff.
So if $(X,\tau)$ is a finite Hausdorff topological space, then the singleton $\{x\}$ is open for each element $x\in X$.  Since $\tau$ includes all arbitrary unions of such singletons, $\tau$ is the power set for $X$, and therefore $(X,\tau)$ is a discrete topological space.
