Assuming countable choice, a generated sigma-algebra is obtained in omega-one stages of complements and countable unions

The displayed successor and limit prescriptions define a class function of the earlier stages, so [L1] produces the unique family $({\mathcal{E}}_{\alpha }{)}_{\alpha <{\omega }_1}$. Each stage is contained in the next because $A$ is the union of the constant sequence with value $A$.

Transfinite induction gives ${\mathcal{E}}_{\alpha }\subseteq {\sigma }_X(\mathcal{E})$ for every $\alpha <{\omega }_1$: at the base, the generated sigma-algebra contains $\mathcal{E}$ and $\mathrm{\varnothing }$; complements and countable unions stay in the sigma-algebra at a successor stage; and a limit stage is a union of earlier subfamilies.

Put $\mathcal{S}:={\bigcup }_{\alpha <{\omega }_1}{\mathcal{E}}_{\alpha }$. It contains $\mathcal{E}$ and $\mathrm{\varnothing }$ and is closed under complements. Given $(A_n)$ in $\mathcal{S}$, [L3] may choose stages ${\alpha }_n$ with $A_n\in {\mathcal{E}}_{{\alpha }_n}$; [L2] bounds the set of chosen stages by some $\beta <{\omega }_1$. Monotonicity from step 1.1 puts every $A_n$ in ${\mathcal{E}}_{\beta }$, so ${\bigcup }_n{A}_n\in {\mathcal{E}}_{\beta +1}\subseteq \mathcal{S}$. Thus $\mathcal{S}$ is a sigma-algebra.

Minimality in [L4] gives ${\sigma }_X(\mathcal{E})\subseteq \mathcal{S}$, while step 2.1 gives the reverse inclusion. Hence the two families are equal.