Assuming the Axiom of Choice, an infinite family E generates at most |E|^aleph-zero sets

Put $\mu :={\kappa }^{{\mathrm{\aleph }}_0}$. Since $2\le \kappa$, [L2] and [L4] give ${\mathrm{\aleph }}_0<2^{{\mathrm{\aleph }}_0}\le \mu$ under [L5]; the minimality in [L6] therefore gives ${\omega }_1\le 2^{{\mathrm{\aleph }}_0}\le \mu$. Also $\mu$ is infinite.

Transfinite induction on the stages in [L1] gives $\mathrm{\mid }{\mathcal{E}}_{\alpha }\mathrm{\mid }\le \mu$. At the base, sending each member of $\mathcal{E}$ to its constant sequence injects $\kappa$ into $\mu$. At a successor, complements contribute at most $\mu$ sets and sequences contribute at most ${\mu }^{{\mathrm{\aleph }}_0}=({\kappa }^{{\mathrm{\aleph }}_0}{)}^{{\mathrm{\aleph }}_0}={\kappa }^{{\mathrm{\aleph }}_0\otimes {\mathrm{\aleph }}_0}=\mu$ by [L2] and [L3]. At a limit below ${\omega }_1$, the predecessor set is countable by [L6], and [L5] chooses stagewise injections into $\mu$; hence the union has size at most ${\mathrm{\aleph }}_0\otimes \mu =\mu$ by [L3].

Using [L5] to choose one injection of each stage into $\mu$, the union of the ${\omega }_1$ stages has cardinal at most ${\omega }_1\otimes \mu =\mu$ by step 1.1 and [L3]. By [L1] this union is ${\sigma }_X(\mathcal{E})$, proving the bound.