Proof verification: Closed under countable union Let $\Omega$ be countable. Let $\{A_i\}_{i \in \mathbb{N}}$ be a countable collection of subsets of $\Omega$. Want to show $\bigcup_{n=1}^\infty A_n \subseteq \Omega$. 
I just want to check the soundness of my argument below: If $x \in \bigcup_{n=1}^\infty A_n $, then $\exists$ $n \in \mathbb{N}$ such that $x \in A_n$. But since $A_n \subseteq \Omega$, we have that $x \in \Omega$. Hence, $\bigcup_{n=1}^\infty A_n \subseteq \Omega$.
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Your proof is correct, although it's worth noting that you didn't use "countable" anywhere. You can be a bit more streamlined by dropping that condition:

Theorem: Let $\Omega$ be a set a $\mathscr U$ be a set of subsets of $\Omega$. Then $\bigcup \mathscr U \subseteq \Omega$.
Proof: Suppose that $x\in \bigcup \mathscr U$. Then, there is some $U\in\mathscr U$ such that $x\in U$. Since $U\subseteq \Omega$ we have $x\in \Omega$. As $x$ was arbitrary, this implies $\bigcup \mathscr U \subseteq \Omega$.

This is pretty much what you wrote verbatim without an indexing set - I'm not suggesting it's any better as a proof (yours is essentially without fault), but it's worth noting when things work in more generality.
