Way Below in Lattice of Power Set

Theorem
Let $X$ be a set.

Let $L = \struct {\powerset X, \cup, \cap, \preceq}$ be a lattice of power set of $X$ where $\mathord\preceq = \mathord\subseteq \cap \paren {\powerset X \times \powerset X}$

Let $x, y \in \powerset X$.

Then $x \ll y$
 * for every a set $Y$ of subsets of $X$ such that $\ds y \subseteq \bigcup Y$
 * then there exists a finite subset $Z$ of $Y$: $\ds x \subseteq \bigcup Z$

where $\ll$ denotes the way below relation.

Sufficient Condition
Let $x \ll y$

Let $Y$ be a set of subsets of $X$ such that
 * $\ds y \subseteq \bigcup Y$

By definitions of power set and subset:
 * $Y \subseteq \powerset X$

By Power Set is Complete Lattice:
 * $\ds \bigcup Y = \sup Y$

By definition of $\preceq$:
 * $y \preceq \sup Y$

By Way Below in Complete Lattice:
 * there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$

By the proof of Power Set is Complete Lattice:
 * $\ds \bigcup Z = \sup Z$

Thus by definition of $\preceq$:
 * there exists a finite subset $Z$ of $Y$: $\ds x \subseteq \bigcup Z$

Necessary Condition
Suppose
 * for every a set $Y$ of subsets of $X$ such that $\ds y \subseteq \bigcup Y$
 * then there exists a finite subset $Z$ of $Y$: $\ds x \subseteq \bigcup Z$

We will prove that
 * for every a subset $Y$ of $\powerset X$ such that $y \preceq \sup Y$
 * then there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$

Let $Y$ be a subset of $\powerset X$ such that
 * $y \preceq \sup Y$

By definition of power set:
 * $Y$ is a set of subsets of $X$.

By Power Set is Complete Lattice:
 * $\ds \bigcup Y = \sup Y$

By definition of $\preceq$:
 * $\ds y \subseteq \bigcup Y$

By assumption:
 * there exists a finite subset $Z$ of $Y$: $\ds x \subseteq \bigcup Z$

By the proof of Power Set is Complete Lattice:
 * $\ds \bigcup Z = \sup Z$

Thus by definition of $\preceq$:
 * there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$

Thus by Way Below in Complete Lattice:
 * $x \ll y$