Noetherian Topological Space is Compact

Theorem
Let $X$ be a Noetherian topological space.

Then $X$ is compact.

Proof
Let $\set{U_i}_{i \in I}$ be a cover of $X$.

That is, $\bigcup_{i \in I}U_i = X$.

Let $V$ be the collection of finite cover of $\set{U_i}_{i\in I}$.

Let $W = \set{\bigcup Y\vert Y \in V}$.

Then $W$ is a collection of open sets.

By Set of Open Sets in Noetherian Space has Maximal Element, $W$ has a maximal element.

Let $\bigcup_{j=1}^n U_{i_j}=U'$ be the maximal element.

Aiming for contradiction, assume that $U' \subsetneq X$.

Let $x \in X \setminus U'$ and let $U_{i_{n+1}}$ be a neighborhood of $x$, where $i_{n+1}\in I$.

Then $U' \cup U_{i_{n+1}}$ is larger than $U'$ and contradicts maximality condition.

Hence $U'$ is a finite subcover.

This shows that $X$ is compact.