Countable union and finite intersection Let $\mathcal{F}$ be a collection of subsets of $\Omega$. Let $\mathcal{F}_{\sigma}$ denote the countable union of sets in $\mathcal{F}$. In the same manner, let $\mathcal{F}_d$ denote the finite intersection of sets in $\mathcal{F}$.
Here is the question: Let's consider the collection $\mathcal{F}_{\sigma d}=(\mathcal{F}_\sigma)_d$. I have to show the collection of sets isn't necessarily closed under countable unions.
I'm looking for a hint. I tried the obvious collection of sets for $\mathcal{F}$: open sets, closed sets, etc..
 A: If I have understood correctly your question, you only want a counterexample showing that is possible to have $\mathcal{F}_{\sigma d}\neq (\mathcal{F}_{\sigma d})_\sigma$.
Basically, this is the reason why, when we try to generate a topology from a collection of subsets, we first have to take closure with respect to finite intersections and then to arbitrary unions; except that we only take countable unions.
Although you asked it in a purely set-theoretic setting, I'll give a geometrically flavoured answear which I think could be more clear, and it's indeed natural if you have seen a bit of topology.
We can take $\Omega=\mathbb{R}$ and $\mathcal{F}$ the union of left and right open halflines, they are respectively of the form $(-\infty,a)$ and $(b,+\infty)$, where $a,b\in\mathbb{R}$.
Thus elements in $\mathcal{F}_\sigma$ are halflines, union of two different types of halflines or even all $\mathbb{R}$.
If we take  $I_1,I_2\in\mathcal{F}_\sigma$ such that $I_i=(-\infty,a_i)\cup 
 (b_i,+\infty)$ where $a_i,b_i\in\mathbb{R}\cup \{-\infty,+\infty\}$ then
$I_1\cap I_2=(-\infty, \min\{a_1,a_2\})\cup (\max\{b_1,b_2\},+\infty)\cup (\min\{b_1,b_2\},\max\{a_1,a_2\})$.
For example, if $a_1=-\infty$, $b_2=+\infty$ and $a_2>b_1$ we get $I_1\cap I_2=(b_1,a_2)$.
Then in $(\mathcal{F}_{\sigma d})_\sigma$ there can be countable disjoint unions of these intervals, like $\bigcup_{n\in\mathbb{N}}(n,n+1)$.
But, iterating intersections, an element in $\mathcal{F}_{\sigma d}$ can only contain a finite number of bounded intervals in the form $(a,b)$ with $a,b\in\mathbb{R}$, so $\mathcal{F}_{\sigma d}\neq (\mathcal{F}_{\sigma d})_\sigma$.
