Are open subsets of Lindelöf spaces themselves Lindelöf? Let $(X,\tau)$ be a Lindelöf space and $O$ an open subset of $X$. Is it true that $O$ is Lindelöf as well?
Recall that a space is Lindelöf if every open cover admits a countable subcover, i.e., this is a weaker notion of compactness.
 A: Consider $\omega_1 + 1$ with the order topology. It is Lindelöf but the open subspace $\omega_1$ is not.
A: The one-point compactification of an uncountable discrete space $X$ is Lindelöf (even compact) but its open subspace $X$ is not Lindelöf.
In fact, every topological space is an open subspace of a compact space.
A: The following are equivalent for any topological space $X$: 


*

*Every open subspace $O$ of $X$ is Lindelöf.

*Every subspace $Y$ of $X$ is Lindelöf.
Proof: clearly 2. implies 1. 
So assume 1., and let $Y$ be any subspace of $X$. To see it is Lindelöf, suppose that $\{U_i, i \in I\}$ is an open cover of $Y$ by relatively open subsets of $Y$. Then for each $i$ pick $O_i$ open in $X$, such that $O_i \cap Y = U_i$.
Define $O = \cup_{i \in I} O_i$, which is open as a union of open sets. 
The $\{O_i: i \in I\}$ form an open cover of $Y$ so 1. implies that there is a countable $J \subset I$ such that $O = \cup_{i \in J} O_i$. But then 
$$ Y = Y \cap O = Y \cap (\bigcup_{i \in J} O_i) = \bigcup_{i \in J} (O_i \cap Y) = \bigcup_{i \in J} U_i\text{,}$$
so the $U_i, i \in J$ form a countable subcover of the original cover and $Y$ is Lindelöf.
So the answer is no, as there are plenty of examples of non-Lindelöf subspaces of Lindelöf spaces.  
