A topological space $X$ is weakly Lindelof if every open cover has a countable subfamily $U$ such that $\bigcup \{ V: V\in U\}$ is dense in X.
Question: Are closed subsets of weakly Lindelof spaces necessarily weakly Lindelof?
A topological space $X$ is weakly Lindelof if every open cover has a countable subfamily $U$ such that $\bigcup \{ V: V\in U\}$ is dense in X. Question: Are closed subsets of weakly Lindelof spaces necessarily weakly Lindelof? 


The Niemytzki plane is weakly Lindelöf (the open upper half plane is a dense Lindelöf subspace); the $x$axis is an uncountable closed and discrete subspace. 


No. Consider the space whose points are all sequences of 0's and 1's of length $\leq\omega$. Visualize it as the binary tree plus "limits" for all paths through the tree, and topologize it accordingly. That is, each finite sequence is an isolated point, but a neighborhood of an infinite sequence $s$ must contain all sufficiently long finite initial segments of $s$. This space is weakly Lindelöf because the finite sequences constitute a countable dense set. But the infinite sequences constitute a closed, discrete, uncountable, and therefore not weakly Lindelöf subspace. 


However, closed subspaces of normal weakly Lindelof spaces are indeed weakly Lindelof. Let $X$ be such a space, $F \subset X$ be closed and $\mathcal{U}$ be an open cover of $F$. If $\mathcal{U}$ covers $X$ then we're done. So we can assume that $G:=X \setminus \bigcup \mathcal{U}$ is a nonempty. Noting that $F$ and $G$ are nonempty disjoint closed sets, use normality to find an open set $O$ such that $G \subset O$ and $\overline{O} \cap F=\emptyset$. Then $\mathcal{U} \cup \{O\}$ is an open cover of the weakly Lindelof space $X$ and hence it contains a countable $\mathcal{V}$ such that $\bigcup \mathcal{V}$ is dense in $X$. Since $\overline{O} \cap F=\emptyset$, the set $\mathcal{C}:=\mathcal{V} \setminus \{O\}$ is a countable subfamily of $\mathcal{U}$ such that $F \subset \overline{\bigcup \mathcal{C}}$. 

