Every Element is Lower implies Union is Lower

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

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

Let
 * $\forall X \in A: X$ is a lower set.

Then $\bigcup A$ is a lower set.

Proof
Let $x \in \bigcup A, y \in S$ such that
 * $y \preceq x$

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

By assumption:
 * $X$ is a lower set.

By definition of lower set:
 * $y \in X$

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