Way Below if Between is Compact Set in Ordered Set of Topology

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $L = \left({\tau, \preceq}\right)$ be an ordered set where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$

Let $x, y \in \tau$ such that
 * $\exists H \subseteq S: x \subseteq H \subseteq y \land H$ is compact

Then
 * $x \ll y$

Proof
Let $D$ be a directed subset of $\tau$ such that
 * $y \preceq \sup D$

By proof of Topology forms Complete Lattice:
 * $y \subseteq \bigcup D$

By Subset Relation is Transitive:
 * $H \subseteq \bigcup D$

By definition:
 * $D$ is open cover of $H$

By definition of compact:
 * $H$ has finite subcover $\mathcal G$ of $D$

By Directed iff Finite Subsets have Upper Bounds:
 * $\exists d \in D: d$ is upper bound for $\mathcal G$

By definitions of upper bound and of $\preceq$:
 * $\forall z \in \mathcal G: z \subseteq d$

By Union is Smallest Superset/Set of Sets:
 * $\bigcup \mathcal G \subseteq d$

By definition of cover:
 * $H \subseteq \bigcup \mathcal G$

By Subset Relation is Transitive:
 * $x \subseteq d$

Thus by definition of $\preceq$:
 * $x \preceq d$

Thus by definition of way below relation:
 * $x \ll y$