Is \bigcap(F \cup G) equal to (\cap F)\bigcap(\cap G)?

[Klungo]

Homework Statement

(Title is wrong)

I was able to prove the similar $\bigcup(F \cup G)=(\cup F)\bigcup(\cup G)$ but I'm not to sure how to go about this one.

Let F and G be nonempty families of sets. Prove
$\bigcap(F \cup G)=(\cap F)\bigcap(\cap G)$

2. The attempt at a solution
To prove $\bigcap(F \cup G) \subset (\cap F)\bigcap(\cap G)$,
let $x \subset \bigcap(F \cup G)$ be arbitrary. Clearly, $x \in S$ for some set $S \subset \bigcap F \cup G$ containing all common elements in $F \cup G$. We have $S \in F \cup G.$

Do I go by..

We have two cases: (only one needs to be proven really.)
Case 1: $If S \in F$, clearly $S \subset \cup F$...
(I'm stuck here)

or Do I go by...

Suppose $S \in F$ and $S \in G$. Clearly $S \subset \cup F$ and $S \subset \cup G$. (I'm stuck here)

I'll try proving the converse and restart the proof since I misread the problem.

 

[voko]

By definition of the operation $\bigcap$, any x in $\bigcap(F\cup G)$ must be contained in EVERY SET of $F\cup G$. Continue from here.