Noetherian Topological Space is Compact/Proof 2

Proof
We may assume $X \ne \O$, since the claim is otherwise trivial.

Let $\CC \subseteq \tau$ be a cover of $X$.

Let:
 * $A := \set { \bigcup \eta : \eta \text{ finite subset of } \CC }$

Observe that $A \ne \O$ and $A \subseteq \tau$,

By, $A$ has a maximal element:
 * $U_1 \cup \cdots \cup U_n$

We shall show that $\set {U_1,\ldots,U_n} \subseteq \CC$ is a finite subcover, i.e.
 * $X = U_1 \cup \cdots \cup U_n$

Otherwise, there would exist:
 * $x \in X \setminus \paren {U_1 \cup \cdots \cup U_n}$

Since $\CC$ is a cover of $X$, there must exist a $V \in \CC$ such that:
 * $x \in V$

But then we would have:
 * $U_1 \cup \cdots \cup U_n \subsetneq U_1 \cup \cdots \cup U_n \cup V$

Since $U_1 \cup \cdots \cup U_n \cup V \in A$, this would contradict the maximality.