Complement of Subset with Property (S) is Closed under Directed Suprema

Theorem
Let $L = \left({S, \preceq}\right)$ be an up-complete ordered set.

Let $X$ be a subset of $S$ with property (S).

Then $\complement_S\left({X}\right)$ is closed under directed suprema.

Proof
Let $D$ be a directed subset of $S$ such that
 * $D \subseteq \complement_S\left({X}\right)$

Aiming for a contradiction suppose that
 * $\sup D \notin \complement_S\left({X}\right)$

By definition of relative complement:
 * $\sup D \in X$

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

By definition of reflexivity:
 * $y \in X$

By definitions of intersection and non-empty:
 * $D \cap X \ne \varnothing$

Thus this by Empty Intersection iff Subset of Complement contradicts
 * $D \subseteq \complement_S\left({X}\right)$