Let $\Omega$ be a countable set. I want to show that every $ \sigma $-algebra $\Sigma$ on $\Omega$ is of the form
\begin{equation} \{ \bigcup_{k \in K'}A_k : K' \subset K \}, \end{equation}
where $A_k$ is a finite or countable partition of $\Omega$.
I already know that for a countable partition $(X_n)_n$ the following holds: \begin{equation} \sigma((A_n)_n) = \{ \bigcup_{n \in J}X_n : J \subset \mathbb{N} \}. \end{equation} So my idea was finding a partition $L$ of $\Omega$ such that is generates the whole $\sigma$-algebra (i.e. $\sigma(L) = \Sigma$). I have found in the following two links what I needed:
Show the sigma algebra of a countable set is generated by a partition
$\sigma$-algebras on a countable set is generated by partitions of the set
So my partition of $\Omega$ is $L:= \{\bigcap_{y \in \Omega}L_{x,y} : x \in \Omega\}$, which was proven to be a partition and described in the above two links.
However, I cannot seem to prove that $\sigma(L) = \Sigma$ holds. I only know for sure that $\sigma(L) \subset \Sigma$, because $\Sigma$ is a $\sigma$-algebra that contains each element of $L$. The other inclusion seems hard to prove, since I have to show that every element $A \in \Sigma$ can be written as a countable union/intersection of these minimal "building blocks". So my question is, in the second link I provided, why can we conclude that this partition generates the whole sigma algebra ($\sigma(L) = \Sigma$)?