Noetherian Topological Space is Compact/Proof 2

Proof
Recall the definition for compact space:

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

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

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

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

$A$ has the following properties:
 * $(1): \quad A \ne \O$
 * $(2): \quad A \subseteq \tau$

Let $\alpha = \bigcup \eta$.

We have:

From $(1)$ and $(2)$, by, $A$ has a maximal element $\alpha$.

We now show that:
 * $\alpha = X$


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

Since $\CC$ is a cover for $X$:
 * $\exists U \in \CC : x \in U$

so that:
 * $\alpha \subsetneqq \alpha \cup U$

This contradicts $(3)$.

Hence by Proof by Contradiction:
 * $\not \exists x \in X \setminus \alpha$

and so:
 * $\alpha = X$

as required.

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

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

This means:
 * $X = U_1 \cup \cdots \cup U_n$

Therefore:
 * $\set {U_1, \ldots, U_n}$ is a finite subcover of $\CC$ for $X$.