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 non-empty. Noting that $F$ and $G$ are non-empty 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 $F \subset \overline{\bigcup \mathcal{C}}$.