PFA implies there are no S-spaces

Assume for contradiction that $X$ is not hereditarily Lindelöf. By [F1] it has a subspace $S$ and an ${\omega }_1$-generated ideal $\mathcal{I}$ of countable subsets of $S$ with the stated witness properties.

By [F2], some uncountable $D\subseteq S$ is inside or outside $\mathcal{I}$. In either case [F1] says that $D$, with its subspace topology, is nonseparable.

But $D$ is also a subspace of $X$, so hereditary separability of $X$ says that $D$ is separable, contradicting step 2.1. Therefore $X$ is hereditarily Lindelöf.

If an S-space existed, [F3] would make it regular Hausdorff and hereditarily separable, so step 3.1 would make it hereditarily Lindelöf and hence Lindelöf. This contradicts the defining non-Lindelöf clause. Thus no S-space exists under PFA. No empty or singleton space can be an S-space because both are Lindelöf; the uncountable witness in step 2.1 is nonempty; and [F4] propagates the exact AC uses of [F1] and [F2].