However, closed subspaces ofif $X$ is a normal weakly Lindelof spaces are indeed weakly Lindelof.
Let $X$ be such a space, $F \subset X$ be$F$ is a closed subspace of $X$ and $\mathcal{U}$ beis an open cover of $F$ then there is a countable subcollection $\mathcal{V}$ of $\mathcal{U}$ such that $F \subseteq \overline{\bigcup \mathcal{V}}$. If
Indeed, 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}$$\mathcal{C}$ such that $\bigcup \mathcal{V}$$\bigcup \mathcal{C}$ is dense in $X$. Since $\overline{O} \cap F=\emptyset$, the set $\mathcal{C}:=\mathcal{V} \setminus \{O\}$$\mathcal{V}:=\mathcal{C} \setminus \{O\}$ is a countable subfamily of $\mathcal{U}$ such that $F \subset \overline{\bigcup \mathcal{C}}$$F \subset \overline{\bigcup \mathcal{V}}$.
This doesn't mean that $F$ is weakly Lindelof though. Indeed one can make the Niemytzki Plane normal by replacing the $x$-axis with a $Q$-set, that is an uncountable subset of the reals whose every subset is a relative $G_\delta$. It is consistent with ZFC that such a set exists. The $Q$-set would still be a closed non-weakly Lindelof subspace of the Niemytzki Plane but, in view of the above, it would at least be "weakly Lindelof" with respect to covers made up of open subsets of $X$.