Hereditarily Lindelof Space In the article "Equilateral sets in Banach spaces of the form $C(K)$" (Mercourakis-Vassiliadis) is stated the following fact:
$$\text{A compact space }K\text{ is Hereditarily Lindelöf}\iff \text{ each closed subset of }K\text{ is }G_{\delta}.$$
I need only the implication $\implies$, but I'm struggling proving it. Can anyone give a either a reference or a proof (at least of $\implies$)?
 A: Let $X$ be hereditarily Lindelöf. If $F$ is closed in $X$, consider $U=F^\complement$, which is open. For each $x \in U$ pick (by $T_3$-ness of $X$) an open neighbourhood $O_x$ of $x$ such that $\overline{O_x} \subseteq U$.
Then $\{O_x: x \in U\}$ is an open cover of $U$, which is Lindelöf by assumption, so there is a countable $N \subseteq U$ such that $$\bigcup\{O_x: x \in N\}=U$$
Because all closures of $O_x$ are also a subset of $U$ we can even say
$$\bigcup\{\overline{O_x}: x \in N\}=U$$
making $U$ an $F_\sigma$ and so by de Morgan
$$F=\bigcap\{\overline{O_x}^\complement: x \in N\}$$
is a $G_\delta$.
Note that we only need regularity of $X$ (which will follow if $X$ is Hausdorff, in your case) for this direction.
For the reverse: to see $X$ is herediatrily Lindelöf, we only need to show all open subsets of $X$ are Lindelöf. As all closed sets are $G_\delta$ by assumption, it follows by de Morgan (as in the previous proof) that all open sets are $F_\sigma$ and thus $\sigma$-compact (as closed sets are compact) and hence Lindelöf (a countable union of finite subcovers is a countable subcover, etc.)
So there we do use the compactness of $X$.
A: Suppose that $K$ is a compact Hausdorff space such that some closed subset $F$ of $K$ is not Gδ. Consider $U = K \setminus F$. Note that $U$ is an open subset of $K$ which is not Fσ. We will show that $U$ is not a Lindelöf subspace of $K$
For each $x \in U$ use regularity to fix an open set $V_x$ such that $x \in V_{x} \subseteq \overline{V_x} \subseteq U$. Clearly $\mathcal{V} = \{ V_{x} \}_{x \in U}$ is an open cover of $U$. Note that if $U_0 \subseteq U$ is such that $\bigcup_{x \in U_0} V_x = U$, then we also have that $\bigcup_{x \in U_0} \overline{V_x} = U$, and since $U$ is not Fσ it cannot be that $U_0$ is countable. Thus there is no countable subcover of $\mathcal{V}$, and so $K$ is not hereditarily Lindelöf.
