TITLE: The Sorgenfrey line is hereditarily Lindelöf
QUESTION [7 upvotes]: How can one show that the Sorgenfrey line is hereditarily Lindelöf (that is, all subspaces of the Sorgenfrey line are Lindelöf)?  
I know the Sorgenfrey line is Lindelöf and hence every closed subspace is Lindelöf.

REPLY [3 votes]: Consider any set $S\subseteq\mathbb R$, and let $\mathcal U$ be a collection of Sorgenfrey-open sets which covers $S$. Let's say that a set of reals is countably covered if it's covered by countably many members of $\mathcal U$. I have to show that $S$ is countably covered.
For $x,y\in\mathbb R$, define the relation $x\equiv y$ to mean that the set $\{s\in S:x\le s\le y\text{ or }y\le s\le x\}$ is countably covered.
It's easy to see that $\equiv$ is an equivalence relation on $\mathbb R$ and that the equivalence classes are convex subsets of $\mathbb R$, i.e., intervals or singletons. Moreover, using the fact that a convex subset of $\mathbb R$ is a countable union of closed intervals, we can see that the intersection of $S$ with each equivalence class is countably covered. Since each point of $S$ is in a nondegenerate equivalence class (an interval, not a singleton), and since there are only countably many such classes, it follows that $S$ is countably covered.