A countable partition generates exactly the unions of its blocks, and the resulting sigma-algebra is countable exactly for a finite partition

Let $\mathcal{U}:=\{{\bigcup }_{i\in S}{P}_i:S\subseteq I\}$. The empty union is empty; the complement of the union indexed by $S$ is the union indexed by $I\setminus S$; and countable unions correspond to unions of the indexing subsets. Thus $\mathcal{U}$ is a sigma-algebra containing every block.

By [L1], ${\sigma }_X(\{P_i:i\in I\})\subseteq \mathcal{U}$. Conversely, every $S\subseteq I$ is at most countable by [L4], so ${\bigcup }_{i\in S}{P}_i$ is a countable union of generators and belongs to the generated sigma-algebra by [L2]. Hence equality holds.

Pairwise disjointness and nonemptiness make $S\mapsto {\bigcup }_{i\in S}{P}_i$ injective, and step 2.1 makes it surjective. For $\mathrm{\mid }I\mathrm{\mid }=k<\mathrm{\infty }$, its domain has $2^k$ members; when $k=0$, the partition is possible exactly for $X=\mathrm{\varnothing }$ and the sigma-algebra is $\{\mathrm{\varnothing }\}$.

If $I$ is countably infinite and the generated sigma-algebra were at most countable, step 3.1 would inject $\mathcal{P}(I)$ into $\mathbb{N}$. Since $I\approx \mathbb{N}$, [L3] would then force $I\approx \mathcal{P}(I)$, contradicting Cantor's theorem. Together with the finite case of step 3.1, this proves both directions of the final equivalence.