Closed subsets of Lindelöf spaces are Lindelöf

Question

Let $X$ be Lindelöf and let $A \subseteq X$ be closed. Show that it follows that $A$ is Lindelöf.

That is, we want to show for every open cover of $A$, there is a countable subcover.

Note: (A space is Lindelöf if every open cover of the space has a countable subcover, a weakening of the better-known notion of compactess).

Answer 1

Using DeMorgan's laws, one can show that a space $X$ is Lindelof iff for every collection $A_\alpha$ of closed subsets of $X$ with the countable intersection property (meaning the intersection of any countable subcollection is nonempty), the total intersection $\cap A_\alpha$ is nonempty. If $A \subset X $ is closed, then every closed set in $A$ is also closed in $X$ so by the previous characterization of Lindelof, $A$ must be Lindelof.

Answer 2

Let $C$ be a cover of $A$. Since $A$ is closed, $A^C$, the complement of $A$, is open. Then $C\cup\bigl\{A^C\bigr\}$ is an open cover of $X$. It contains a countable subcover $C'$. Then $C'\setminus\bigl\{A^C\bigr\}$ is countable, covers $A$, and is a subcover of $C$.