TITLE: Artinian topological space are compact
QUESTION [5 upvotes]: Call a topological space $X$ Artinian if every nested sequence of closed sets 
$$C_1 \supset C_2 \supset C_3 \supset \cdots$$
is eventually constant.

Prove that if $X$ is Artinian then it is also compact.

I don't have a concrete strategy for attacking this. I am assuming (maybe wrongly so) that appealing to the finite intersection property might be of some help. Is this the case? Should I employ a different strategy? 
Thank you very much in advance!

REPLY [5 votes]: Take an open cover $\{U_i\mid i\in I\}$. Using the axiom of choice we can assume that $I$ is well-ordered and so we can write $V_i=\bigcup_{j<i}U_i$, then $\{V_i\mid i\in I\}$ is an open cover as well.
Use the usual trick of complements and the Artinian property to conclude there is a finite subcover, or else.