Set of Upper Closures of Compact Elements is Basis implies Complete Scott Topological Lattice is Algebraic

Theorem
Let $L = \left({S, \preceq, \tau}\right)$ be a complete Scott Definition:Topological Lattice.

Let $\mathcal B = \left\{ {x^\succeq: x \in K\left({L}\right)}\right\}$ be a basis of $L$

where $K\left({L}\right)$ denotes the compact subset of $L$.

Then $L$ is algebraic.

Proof
Thus by Compact Closure is Directed:
 * $\forall x \in S:x^{\mathrm{compact} }$ is directed

where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.

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

Let $x \in S$.

By definition of lower closure of element:
 * $x$ is upper closure for $x^\preceq$

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

By Compact Closure is Intersection of Lower Closure and Compact Subset:
 * $x^{\mathrm{compact} } = x^\preceq \cap K\left({L}\right)$

By Intersection is Subset:
 * $x^{\mathrm{compact} } \subseteq x^\preceq$

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

By definition of transitivity:
 * $\sup \left({x^{\mathrm{compact} } }\right) \preceq x$

Aiming for a contradiction suppose that
 * $x \ne \sup \left({x^{\mathrm{compact} } }\right)$

We will prove that
 * $x \notin \left({\left({x^{\mathrm{compact} } }\right) }\right)^\preceq$

Aiming for a contradiction suppose that
 * $x \in \left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq$