Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice.

Let $x, y \in S$ such that
 * $x \ll y$

where $\ll$ denotes the way below relation.

Then there exists a way below open filter in $L$: $y \in F \land F \subseteq x^\gg$

where $x^\gg$ denotes the way above closure of $x$.