Noetherian Topological Space is Compact/Proof 2

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

We shall verify.

Let $\CC \subseteq \tau$ be an arbitrary cover of $X$.

We shall show that $\CC$ has a finite subcover.

Consider:
 * $A := \leftset {\bigcup \eta: \eta}$ is a finite subset of $\rightset \CC$

The following trivial properties are trivial:
 * $\paren 1$: $\CC \subseteq A$.
 * $\paren 2$: $A \ne \O$.
 * $\paren 3$: $A \subseteq \tau$.
 * $\paren 4$: $\alpha, \beta \in A \implies \alpha \cup \beta \in A$.
 * $\paren 5$: $\alpha \in A, U \in \CC \implies \alpha \cup U \in A$.

From $\paren 2$ and $\paren 3$, by, $A$ has a maximal element $\alpha$.

We claim that:
 * $\paren \star \quad \alpha = X$


 * $\exists x \in X \setminus \alpha$
 * $\exists x \in X \setminus \alpha$

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

Then, by $\paren 5$, we have:
 * $\alpha \cup U \in A$

This contradicts the maximality of $\alpha$, since:
 * $\alpha \subsetneq \alpha \cup U$.

Finally, since $\alpha \in A$, we can write it as:
 * $\alpha = U_1 \cup \cdots \cup U_n$

using $U_1,\ldots,U_n \in \CC$ for some $n \in \N_{>0}$.

By $\paren \star$, this means:
 * $X = U_1 \cup \cdots \cup U_n$

This is nothing but:
 * $\set {U_1,\ldots,U_n} \subseteq \CC$ is a finite subcover.