TITLE: A finite Hausdorff space is discrete
QUESTION [13 upvotes]: 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?

REPLY [4 votes]: 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 $T_1$.