Let $\Omega$ be a set $(A_n)_{n \in \mathbb{N}}$ a sequence of subsets of $\Omega$ such that $A_m \cap A_n = \emptyset$ whenever $m \neq n$ and $\bigcup_{n\in \mathbb{N}}A_n = \Omega$. Consider the family $\mathcal{A}$ of subsets of $\Omega$ defined by:
$\mathcal{A}:= \left\{\bigcup_{i \in I} A_i : I \subseteq \mathbb{N} \right\}$.
To show: $\mathcal{A}$ is a $\sigma$-algebra.
My attempt:
Choose $I = \mathbb{N}$ then $\bigcup_{i \in \mathbb{N}} A_i \in \mathcal{A}$ and by assumption $\bigcup_{i \in \mathbb{N}} A_i = \Omega$
Let $A\in \mathcal{A}$ then $A=\bigcup_{i \in \lambda} A_i $ for $\lambda \subset \mathbb{N}$. Now $A^c = (\bigcup_{i \in \lambda} A_i)^c = \bigcup_{j\in \mathbb{N}\setminus \lambda}A_j $. This is true because $A_m \cap A_n = \emptyset$ whenever $m \neq n$. But since ($\mathbb{N} \setminus \lambda) \subset \mathbb{N} \implies A^c \in \mathcal{A}$
Let $A_{I_n} \in \mathcal{A}$. Then $A_{I_n} = \bigcup_{i \in I_n}A_i.$ But then $\bigcup_{n \in \mathbb{N}} A_{I_n} = \bigcup_{i \in \mathbb{N}}A_i$ which is by assumption in $\mathcal{A}$.
Is that correct?
Since your proposed sigma algebra consists of the various unions of a specific list of pairwise disjoint sets, it is obvious anyway that if you take a countable union of those it will be in your proposed sigma algebra, merely because that proposed algebra was already defined to be the collection of all subsets of the power set of the basic collection of pairwise disjoint sets.
It still seems there might be a simpler way to say the above; the more I look at it my version seems overly complicated.