Upper Closure is Compact in Topological Lattice

Theorem
Let $L = \left({S, \preceq, \tau}\right)$ be a topological lattice.

Suppose that
 * for every subset $X$ of $S$ if $X$ is open, then $X$ is upper.

Let $x \in S$.

Then $x^\succeq$ is compact

where $x^\succeq$ denotes the upper closure of $x$.

Proof
Let $\mathcal F$ be a set of subsets of $S$ such that:
 * $\mathcal F$ is open cover of $x^\succeq$

By definition of cover:
 * $x^\succeq \subseteq \bigcup \mathcal F$

By definitions of upper closure of element and reflexivity:
 * $x \in x^\succeq$

By definition of subset:
 * $x \in \bigcup \mathcal F$

By definition of union:
 * $\exists Y \in \mathcal F: x \in Y$

Defime $\mathcal G = \left\{ {Y}\right\}$

By definition of open cover:
 * $Y$ is open.

We will prove that
 * $x^\succeq \subseteq \bigcup \mathcal G$

Let $y \in x^\succeq$

By definition of upper closure of element:
 * $x \preceq y$

By Union of Singleton:
 * $\bigcup \mathcal G = Y$

By assumption:
 * $Y$ is upper.

Thus by definition of upper set:
 * $x \in \bigcup \mathcal G$

Then by definition:
 * $\mathcal G$ is cover of $x^\succeq$

By definitions of singleton and subset:
 * $\mathcal G \subseteq \mathcal F$

By definition:
 * $\mathcal G$ is subcover of $\mathcal F$.

By Singleton is Finite:
 * $\mathcal G$ is finite.

Thus by definition:
 * $\mathcal G$ is finite subcover of $\mathcal F$.