Relational Structure with Topology of Subsets with Property (S) is Topological Space

Theorem
Let $T = \left({S, \preceq, \tau}\right)$ be a relational structure with topology

where
 * $\left({S, preceq}\right)$ is an up-complete ordered set,
 * $\tau$ is the set of all subsets of $S$ with property (S).

Then $\left({S, \tau}\right)$ is topological space.

Proof
We will prove that
 * $S$ has property (S).

Let $D$ be a directed subset of $S$ such that
 * $\sup D \in S$

By definition of non-empty set:
 * $\exists y: y \in D$

Thus $y \in D$.

Thus by definition of subset:
 * $\forall x \in D: y \preceq x \implies x \in S$

Then $(O3): \quad S \in \tau$

We will prove that
 * $(O1): \quad \forall F \subseteq \tau: \bigcup F \in \tau$

Let $F \subseteq \tau$.

To prove that $\bigcup F \in \tau$ it should be proved that
 * $\bigcup F$ has property (S).

Let $D$ be a directed subset of $S$ such that
 * $\sup D \in \bigcup F$

By definition of union:
 * $\exists X \in F: \sup D \in X$

By definition of subset:
 * $X \in \tau$

By assumption:
 * $X$ has property (S).

By definition of property (S):
 * $\exists y \in D: \forall x \in D: y \preceq x \implies x \in X$

Thus $y \in D$.

Let $x \in D$ such that
 * $y \preceq x$

Then $x \in X$.

Thus by definition of union:
 * $x \in \bigcup F$

We will prove that
 * $(O2): \quad \forall X, Y \in \tau: X \cap Y \in \tau$

Let $X, Y \in \tau$.

By assumption:
 * $X$ and $Y$ have property (S).

To prove that $X \cap Y \in \tau$ it should be proved that
 * $X \cap Y$ has property (S).

Let $D$ be a directed subset of $S$ such that
 * $\sup D \in X \cap Y$

By definition of intersection:
 * $\sup D \in X$ and $\sup D \in Y$

By definition of property (S):
 * $\exists x \in D: \forall z \in D: x \preceq z \implies z \in X$

and
 * $\exists y \in D: \forall z \in D: y \preceq z \implies z \in X$

By definition of directed:
 * $\exists z \in D: x \preceq z \land y \preceq z$

Thus $z \in D$.

Let $u \in D$ such that
 * $z \preceq u$

By definition of transitivity:
 * $x \preceq u$ and $y \preceq u$

Then
 * $u \in X$ and $u \in Y$

Thus by definition of intersection:
 * $u \in X \cap Y$

Hence by $(O1)$, $(O2)$, $(O3)$, and definition:
 * $\left({S, \tau}\right)$ is topological space.