Please Help me understand this proof 

DOUBT
What i didnot understand is from where it is written our new goal means there exists a... I didnot understand how there exists word popped up here and why the new givens are written as they are, i mean using there exist and for all
Thanks
 A: There is a lot of blabla in the excerpt shown. Before you are going into the finicky details of the logic atoms used in this proof you should understand it on a real world level. 
We are given two families of sets ${\cal F}$ and ${\cal G}$, and are told that ${\cal F}\cap{\cal G}\ne\emptyset$. This means that there is at least one set $A$ belonging to both families.
We next consider the set $\cap{\cal F}$, which is shorthand for the intersection of all sets $X_\alpha\in{\cal F}$, and similarly $\cup{\cal G}$, which is shorthand for the union of all sets $Y_\beta\in {\cal G}$.
Claim:  $\quad{\cal F}\cap{\cal G}\ne\emptyset\quad\Longrightarrow\quad \cap{\cal F}\ \subset\ \cup{\cal G}$.
Proof: Any $x\in\cap{\cal F}$ belongs to every $X_\alpha\in{\cal F}$, in particular to the set $A\in{\cal F}\cap{\cal G}$ mentioned above. This set $A$ takes also part in the grand union $\cup{\cal G}$; therefore $x$ belongs to this union.$\quad\square$
Note that the proof took only two lines.
A: We don't need to "choose" A. We know that at least one set A belongs to F and to G. The intersection of F is a subset of any such A, because A belongs to F, and any such A is a subset of union-G, because A belongs to G. QED. 
A: This should be enough as a proof:
Let $\mathcal H=\mathcal F\cap\mathcal G$. As $\;\mathcal H\subset \mathcal F$, we have $\;\bigcap\mathcal F\subset\bigcap\mathcal H $.
Also $\;\bigcap \mathcal H\subset\bigcup \mathcal H $. 
Finally, as  $\;\mathcal H\subset \mathcal G$, $\;\bigcup\mathcal H\subset \bigcup\mathcal G$. Then use transitivity of inclusion.
