Every Element is Directed and Every Two Elements are Included in Third Element implies Union is Directed

Theorem
Let $P = \left({S, \preceq}\right)$ be an ordered set.

Let $A$ be a set of subsets of $S$.

Let
 * $\forall X \in A: X$ is directed.

Let
 * $\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$

Then $\bigcup A$ is directed.

Proof
Let $x, y \in \bigcup A$.

By definition of union:
 * $\exists X \in A: x \in X$

and
 * $\exists Y \in A: y \in Y$

By assumption:
 * $\exists Z \in A: X \cup Y \subseteq Z$

By definition of union:
 * $x, y \in X \cup Y$

By definition of subset:
 * $x, y \in Z$

By assumption:
 * $Z$ is directed.

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

Thus by definition of union:
 * $z \in \bigcup A$

Thus
 * $x \preceq z \land y \preceq z$

Hence $\bigcup A$ is directed.