Finite T1 Space is Discrete/Proof 1

Proof
Let $T = \left({S, \tau}\right)$ be a $T_1$ space on a finite set $S$.

Let $U \subseteq S$ be any subset of $S$.

Let $H = \complement_S \left({U}\right)$ be the complement of $U$ relative to $S$.

Then by Relative Complement of Relative Complement we have that $U = \complement_S \left({H}\right)$.

We can write $H$ as:
 * $\displaystyle H = \bigcup_{x \in H} \left\{{x}\right\}$

From Equivalence of Definitions of $T_1$ Space, $\forall x \in H: \left\{{x}\right\}$ is closed in $T$.

As $S$ is a finite set, it follows that $H$ is the union of a finite number of closed sets of $T$.

By Topology Defined by Closed Sets, $H$ is therefore closed in $T$.

By definition of closed set, $U = \complement_S \left({H}\right)$ is open in $T$.

As $U$ is arbitrary, it follows that:
 * $\forall U \subseteq S: S \in \tau$

by definition of a topology.

Hence the result, by definition of the discrete topology.