Element is Finite iff Element is Compact in Lattice of Power Set

Theorem
Let $X$ be a set.

Let $L = \left({\mathcal P\left({X}\right), \cup, \cap, \subseteq}\right)$ be a lattice of power set.

Let $x \in \mathcal P\left({X}\right)$.

Then $x$ is a finite set $x$ is a compact element.

Proof
The case when $x = \varnothing$

By Empty is Bottom of Lattice of Power Set:
 * $x = \bot$

where $\bot$ denotes the bottom of $L$.

By Bottom is Way Below Any Element:
 * $x \ll x$

where $\ll$ denotes the way below relation.

Thus by definition
 * $x$ is a finite set $x$ is a compact element.

The case when $x \ne \varnothing$:

Sufficient Condition when Non-empty
Let $x$ be a finite set.

We will prove that
 * for every a set $Y$ of subsets of $X$ such that $x \subseteq \bigcup Y$
 * then there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$

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

By definitions of union and subset:
 * $\forall e \in x: \exists u \in Y: e \in u$

By Axiom of Choice:
 * $\exists f:x \to Y: \forall e \in x: e \in f\left({e}\right)$

Define $Z = f\left[{x}\right]$

By Image of Mapping from Finite Set is Finite:
 * $Z$ is a finite set.

By definition of image of mapping:
 * $Z \subseteq Y$

It remains to prove that
 * $x \subseteq \bigcup Z$

Let $e \in x$.

By definitions of $f$ and image of mapping:
 * $e \in f\left({e}\right) \in Z$

Thus by definition of union:
 * $e \in \bigcup Z$

By Way Below in Lattice of Power Set:
 * $x \ll x$

Thus by definition of compact:
 * $x$ is compact.

=== Necessary Condition