Infinite Subset of Finite Complement Space Intersects Open Sets

Theorem
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.

Let $H \subseteq S$ be an infinite subset of $S$.

Then the intersection of $H$ with any non-empty open set of $T$ is infinite.

Proof
Let $U \in \tau$ be any non-empty open set of $T$.

Then by definition of finite complement topology:
 * $\relcomp S U$ is finite.

We have that:
 * $H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$

$H \cap U$ is finite.

From Intersection is Subset:
 * $H \cap \relcomp S U \subseteq \relcomp S U$

From Subset of Finite Set is Finite:
 * $H \cap \relcomp S U$ is also finite.

Thus we have:
 * $H = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$

which is the union of two finite sets.

From Union of Finite Sets is Finite, $H$ is finite.

It is a contradiction that $H$ is infinite and finite at the same time.

Hence by Proof by Contradiction, $H \cap U$ must be infinite.