Element equals to Supremum of Infima of Open Sets that Element Belongs implies Topological Lattice is Continuous

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

Let
 * $\forall x \in S: x = \sup \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$

Then $L$ is continuous.

Proof
Thus by Way Below Closure is Directed in Bounded Below Join Semilattice:
 * $\forall x \in S:x^\ll$ is directed.

Thus by definition pf complete lattice:
 * $L$ is up-complete.

Let $x \in S$.

Define $W := \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$

By definition of way below closure:
 * $x$ is upper bound for $x^\ll$

By definition of supremum:
 * $\sup \left({x^\ll}\right) \preceq x$

We will prove that
 * $W \subseteq x^\ll$

Let $d \in W$.

By definition of $W$:
 * $\exists V: d = \inf V \land x \in V \land V \in \sigma\left({L}\right)$

By Scott Topology equals to Scott Sigma:
 * $\tau = \sigma\left({L}\right)$

By definition:
 * $V$ is open.

By Infimum of Open Set is Way Below Element in Complete Scott Topological Lattice:
 * $d \ll x$

Thus by definition of way below closure:
 * $d \in x^\ll$

By Supremum of Subset:
 * $\sup W \preceq \sup\left({x^\ll}\right)$

By assumption:
 * $x \preceq \sup\left({x^\ll}\right)$

Hence by definition of antisymmetry:
 * $x = \sup \left({x^\ll}\right)$