Complement of Irreducible Topological Subset is Prime Element

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $X$ be an irreducible subset of $S$ such that
 * $\complement_S\left({X}\right) \in \tau$

Let $L = \left({\tau, \preceq}\right)$ be an inclusion ordered set of topology $\tau$.

Then $\complement_S\left({X}\right)$ is prime element in $L$.

Proof
Let $Y, Z \in \tau$ such that
 * $Y \wedge Z \preceq \complement_S\left({X}\right)$

By definition of topological space:
 * $Y \cap Z \in \tau$

By Meet in Inclusion Ordered Set:
 * $Y \cap Z = Y \wedge Z$

By definition of inclusion ordered set:
 * $Y \cap Z \subseteq \complement_S\left({X}\right)$

By Relative Complement is Decreasing and Relative Complement of Relative Complement:
 * $X \subseteq \complement_S\left({Y \cap Z}\right)$

By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection:
 * $X \subseteq \complement_S\left({Y}\right) \cup \complement_S\left({Z}\right)$

By Intersection with Subset is Subset:
 * $X = \left({\complement_S\left({Y}\right) \cup \complement_S\left({Z}\right)}\right) \cap X$

By Intersection Distributes over Union:
 * $X = \left({\complement_S\left({Y}\right) \cap X}\right) \cup \left({\complement_S\left({Z}\right) \cap X}\right)$

By definition of closed set and Relative Complement of Relative Complement:
 * $X$, $\complement_S\left({Y}\right)$, and $\complement_S\left({Z}\right)$ are closed sets.

By Intersection of Closed Sets is Closed in Topological Space:
 * $\complement_S\left({Y}\right) \cap X$, $\complement_S\left({Z}\right) \cap X$ are closed sets.

By definition of irreducible:
 * $\complement_S\left({Y}\right) \cap X = X$ or $\complement_S\left({Z}\right) \cap X = X$

By Intersection with Subset is Subset:
 * $X \subseteq \complement_S\left({Y}\right)$ or $X \subseteq \complement_S\left({Z}\right)$

By Relative Complement is Decreasing and Relative Complement of Relative Complement:
 * $Y \subseteq \complement_S\left({X}\right)$ or $Z \subseteq \complement_S\left({X}\right)$

Thus by definition of inclusion ordered set:
 * $Y \preceq \complement_S\left({X}\right)$ or $Z \preceq \complement_S\left({X}\right)$