A countable generator of a sigma-algebra yields a countable algebra of sets

If $\mathcal{G}=\mathrm{\varnothing }$, then $\{\mathrm{\varnothing },X\}$ is a finite algebra [L1, given, algebra] of subsets of $X$ containing $\mathcal{G}$, so the conclusion holds. Assume now that $\mathcal{G}\mathrm{\ne }\mathrm{\varnothing }$.

Enumerate $\mathcal{G}=\{G_1,G_2,\dots \}$. For each $m\ge 1$, let [L1, given, choose, construct] ${\mathcal{A}}_m$ be the family of all unions of atoms of the finite partition generated by $G_1,\dots ,G_m$; equivalently, the members of ${\mathcal{A}}_m$ are all finite Boolean combinations of those $m$ sets. Each ${\mathcal{A}}_m$ is a finite algebra on $X$ containing $G_1,\dots ,G_m$.

Put [step 1.2, L1, choose, algebra] $\mathcal{A}:={\bigcup }_{m\ge 1}{\mathcal{A}}_m\mathrm{.}$ Then $\mathcal{A}$ contains every $G_j$. If $A,B\in \mathcal{A}$, choose $m$ with $A,B\in {\mathcal{A}}_m$; because ${\mathcal{A}}_m$ is an algebra, also $X\setminus A$ and $A\cup B$ lie in ${\mathcal{A}}_m\subseteq \mathcal{A}$. Thus $\mathcal{A}$ is an algebra of subsets of $X$.

Each ${\mathcal{A}}_m$ is finite, so in particular countable, and [L2] makes [L2, step 2.1] their countable union $\mathcal{A}$ countable. Therefore $\mathcal{A}$ is a countable algebra containing $\mathcal{G}$.