Union of Filtered Sets is Filtered

Theorem
Let $\left({S, \preceq}\right)$ be a preordered set.

Let $A$ be a set of subsets of $S$ such that
 * $\forall X \in A: X$ is filtered

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

Then:
 * $\bigcup A$ is also filtered.

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 filtered.

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

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

Hence
 * $\bigcup A$ is filtered.