An easy (or not?) collection of proper sets . Let $S$ be a finite set.
We are given $k$ rows and in each row we have two subsets of $S$ which we call them  $A_i$, $B_i$ (for the $i$th row, with  $i\leq k$).  
$A_1$ and $B_1$
$A_2$ and $B_2$
.
.
.
$A_k$ and $B_k$   
with two properties satisfied  :
1.$A_i\cap B_i=\emptyset$
2.For every two elements of $S$ (let's call them $p,q$ for convience) the following holds:
If $\{p, q\}\subseteq (A_i\cup B_i)$  then 
$\{p,q\}\not\subseteq(A_j\cup B_j),i\neq j$
(in other words: for every $p$ and $q$ from $S$, both $p$ and $q$ cannot appear together in two rows)
I want to prove (or disprove but I doubt it) that you can choose a set from every row in such a way to have  a collection of $k$ disjoint sets.   
Example.
Let $S=\{a,b,c,d,e,f\}$
We can see that the two conditions are satisfied in these rows:
$\{e,a \}$ and $\{b,c \}$
$\{c \}$ and  $\{d \}$
$\{f,a \}$ and  $\{d \}$
$\{d \}$ and $\{b \}$
So, we can choose from each row the disjoint sets $\{e,a \} ,\{c \} ,\{d \} , \{b \}$ respectively.   
This problem has puzzled me a lot and I would really appreciate if somebody could help me here.    
EDIT as Elaqqad has mentioned,we must assume also that $|S|\geq k$ otherwise the problem has a direct  solution as it is mentioned below. 
Thanks in advance!
 A: This is not True in genral ; let's take $S=\{a,b,c,d\}$ and let's take our rows:
$\{a \}$ and  $\{b \}$ 
$\{b \}$ and  $\{c \}$ 
$\{c \}$ and  $\{d \}$ 
$\{d \}$ and  $\{a \}$ 
Until here every thing works, this example verifies your assumptions, and because the subsets are all singletons then it verifies the second assumption if and only if we did not repeat any row. Now the idea is that we must choose $k=4$ subsets from the rows which are all disjoints, this is possible but again because we have singletons we must have all elements of $\{a,b,c,d\}$ present in the union of the chosen collection. And the counter example  comes when we add another row :
$\{ a \}$ and $\{c\}$
By adding this row, as we know we have to verify that the assumptions are false, otherwise we must choose $5$ singletons which will never be disjoints (because the initial set contains only $4$ elements). But the your assumptions are also true because there is no repeated row (as we explained in the first paragraph).
Edit
Even the assumption $k\leq |S|$ is not sufficient. I don't really see any relation between the conditions and the result you want to prove, Let's take $S=\{a,b,c,d,e,f,g,h\}$ and consider the following rows:
$\{a,b\}$ and  $\{c,d \}$ 
$\{e,f\}$ and  $\{g,h\}$ 
$\{a \}$ and  $\{e \}$ 
$\{b \}$ and  $\{f \}$ 
$\{c \}$ and  $\{g \}$ 
$\{d \}$ and  $\{h \}$ 
And p to here everything works, but we can add another row :
$\{a \}$ and  $\{h \}$ 
and this is a counter example for $k=7$ and $|S|=8$
