Let $\emptyset \not = A \subset P(\omega)$ , find a set $B \subset A$ such that :
1) $B$ is countable
2) $\bigcap B = \bigcap A$
3) $\bigcup B = \bigcup A$
If $B$ was a set that is not necessarily subset of $A$ then i could just take $B = \{0,1,2,3,\cdots\} \cup \{\omega\}$
Easy to show that $B$ is countable and then reduce $B$ till we have that $\bigcap B= \bigcap A$ and $\bigcup B = \bigcup A$
So this condition $B \subset A$ that is making the proof a problem for me.
You need to use the axiom of choice here, since otherwise it is consistent that there is a set $A$ which is infinite, but has no uncountably infinite subset.
But let's try to break this down to just one problem. Let's say that you only wanted to have a countable subset $B$ such that $\bigcup A=\bigcup B$. How would you go about doing that? Well, for every $n\in\bigcup A$, choose some $A_n\in A$ such that $n\in A$. Now take $B=\{A_n\mid n\in\bigcup A\}$, and we immediately have that:
The last one is true because every $n$ in that union was forced into the union of $B$ by some witness that it is at all in the union of $A$.
Now think about the way this works, try to get the same proof with $\bigcap$ instead of $\bigcup$, and then solving your problem should be visible.