CH implies that an S-space exists

Let $D\subseteq \mathcal{C}$ consist of the binary sequences of finite support. Sending a finite support $a$ to ${\sum }_{k\in a}{2}^k$ enumerates $D$ bijectively by $\omega$, and $D$ meets every cylinder: extend the prescribed finite word by zeros. The map $x\mapsto y_x$, where $y_x(2k)=x(k)$ and $y_x(2k+1)=1$, injects $\mathcal{C}$ into $\mathcal{C}\setminus D$. Inclusion gives the reverse injection, so Cantor--Bernstein and [F2] give $\mathrm{\mid }\mathcal{C}\setminus D\mathrm{\mid }=\mathrm{\mid }\mathcal{C}\mathrm{\mid }={\mathrm{\aleph }}_1$.

Choose a bijection $b:{\omega }_1\to \mathcal{C}$ with $b[\omega ]=D$: use the enumeration from step 1.1 on $\omega$ and a bijection ${\omega }_1\setminus \omega \to \mathcal{C}\setminus D$ on the complements. Pull the cylinder topology back along $b$. It is Hausdorff, zero-dimensional, separable, and second countable, with $\omega$ dense. Every nonempty open set contains a cylinder. For a word $s$, prefixing $s$ defines a bijection from $\mathcal{C}$ onto its cylinder $N_s$, so every nonempty open set has size ${\mathrm{\aleph }}_1$.

For $\alpha <{\omega }_1$ and $n<\omega$, put $W_n^{\alpha }=b^{-1}[N_{b(\alpha )\upharpoonright n}]$. Then $W_0^{\alpha }={\omega }_1$, each $W_n^{\alpha }$ is clopen, and these sets form a local base at $\alpha$. They are strictly decreasing because the next unrestricted bit can be changed, and their intersection is $\{\alpha \}$. Consequently step 2.1 with these bases is a second-countable ordered fundamental space in the exact sense of [F3].

Apply [F4] to obtain a nice refinement $X=({\omega }_1,\tilde{\mathcal{T}})$ satisfying every $({\star }_{\alpha })$. The space $X$ is regular and Hausdorff and is not Lindelöf.

Fix $0<n<\omega$. By [F5], $X^n$ is hereditarily separable. By [F6], $X^n$ is regular and Hausdorff.

The power $X^n$ is not Lindelöf. Otherwise let $\mathcal{U}$ be an open cover of $X$ with no countable subcover, supplied by step 4.1. The inverse images $\{{\pi }_0^{-1}[U]:U\in \mathcal{U}\}$ form an open cover of $X^n$. Lindelöfness would give countably many of them covering $X^n$. The projection ${\pi }_0:X^n\to X$ is onto: fill all coordinates other than $0$ with the fixed point $0\in {\omega }_1$. Hence the corresponding countable members of $\mathcal{U}$ would cover $X$, a contradiction.

Steps 5.1 and 5.2 show that every positive finite power of $X$ is regular, Hausdorff, hereditarily separable, and not Lindelöf. Thus every such power is an S-space, and [F7] says exactly that $X$ is a strong S-space. The case $n=1$ is included, while $n=0$ is deliberately excluded. The construction is nonempty because its underlying set is ${\omega }_1$. Step 2.1 is the only new choice in this assembly; [F8] also propagates the ZFC choices in the two refinement suppliers.