Compact Closure is Set of Finite Subsets in Lattice of Power Set

Theorem
Let $X$ be a set.

Let $L = \left({\mathcal P\left({X}\right), \cup, \cap, \preceq}\right)$ be the lattice of power set of $X$ where $\mathord\preceq = \mathord\subseteq \cap \mathcal P\left({X}\right) \times \mathcal P\left({X}\right)$

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

Then $x^{\mathrm{compact} } = \mathit{Fin}\left({x}\right)$

where $\mathit{Fin}\left({x}\right)$ denotes the set of all finite subsets of $x$.

$\subseteq$
Let $y \in x^{\mathrm{compact} }$.

By definition of compact closure:
 * $y \preceq x$ and $y$ is compact.

By definition of $\preceq$:
 * $y \subseteq x$

By Element is Finite iff Element is Compact in Lattice of Power Set: "$y$ is a finite set.

Thus by definition of $\mathit{Fin}$:
 * $y \in \mathit{Fin}\left({x}\right)$

$\supseteq$
Let $y \in \mathit{Fin}\left({x}\right)$.

By definition of $\mathit{Fin}$:
 * $y \subseteq x$ and $y$ is finite.

By definition of $\preceq$:
 * $y \preceq x$

By Element is Finite iff Element is Compact in Lattice of Power Set:
 * $y$ is compact.

Thus by definition of compact closure:
 * $y \in x^{\mathrm{compact} }$