Moore's topology is hereditarily Lindelof

Suppose some subspace $Z\subseteq X$ has an open cover $\mathcal{U}$ with no countable subcover. Recursively for $\alpha <{\omega }_1$, after choosing $U_{\xi }\in \mathcal{U}$ for $\xi <\alpha$, choose $x_{\alpha }\in Z\setminus {\bigcup }_{\xi <\alpha }{U}_{\xi }$ and then $U_{\alpha }\in \mathcal{U}$ containing $x_{\alpha }$. The uncovered remainder is uncountable at every stage: if it were countable, one further cover member for each remaining point, together with the previous countable family, would be a countable subcover. Hence choose $x_{\alpha }$ above all earlier $x_{\xi }$. Thus $⟨x_{\alpha }⟩$ is strictly increasing and $U_{\alpha }$ contains no $x_{\beta }$ for $\beta >\alpha$.

By [F2], shrink each $U_{\alpha }$ around $x_{\alpha }$ to a finite Boolean basic set $V_{\alpha }$ in the subspace $Z$. Intersect also with $W_{x_{\alpha }}$, so its finite support $F_{\alpha }\subseteq X$ contains $x_{\alpha }$. Apply [F3] and the finite pigeonhole principle to retain an uncountable index set on which the supports form a $\mathrm{\Delta }$-system with root $F$, have fixed root and petal positions and one fixed membership-bit string, and have nonempty petals of one size $k$. Pass to a tail so every root member is below every retained $x_{\alpha }$.

Let $A=\{F_{\alpha }\setminus F:\alpha \text{ retained}\}$ and $B=\{\{x_{\beta }\}:\beta \text{ retained}\}$. The family $A$ is uncountable and pairwise disjoint by the $\mathrm{\Delta }$-system property; $B$ is uncountable and pairwise disjoint because the sequence is strictly increasing. Map each petal coordinate to the sole column and prescribe the corresponding fixed membership bit of $V_{\alpha }$. By [F4], choose a petal $F_{\alpha }\setminus F$ and a singleton $\{x_{\beta }\}$ with the petal below $x_{\beta }$ realizing all those bits.

Since $x_{\alpha }$ belongs to its petal, the inequality from step 3.1 gives $x_{\alpha }<x_{\beta }$, and strict increase gives $\alpha <\beta$. The realized petal bits say that $x_{\beta }$ meets every petal condition defining $V_{\alpha }$. For a root coordinate $\eta$, the point $x_{\beta }$ meets its required bit because $x_{\beta }\in V_{\beta }$, the root bit string is uniform, and $\eta <x_{\beta }$ lets [F2] read membership through $c(\eta ,x_{\beta })$. Consequently $x_{\beta }\in V_{\alpha }\subseteq U_{\alpha }$.

Step 1.1 says that $U_{\alpha }$ contains no $x_{\beta }$ with $\beta >\alpha$, contradicting step 4.1. Therefore every subspace $Z$ is Lindelöf, which is precisely hereditary Lindelöfness by [F1]. Empty and countable $Z$ cause no problem: a countable space has a countable subcover by choosing one cover member per point, and the empty space uses the empty subcover.