Let $A_1, A_2, ..., A_r$ be a sequence of subsets(not neccesarily distinct) of an $n$-element set $X$ s.t. each set has size $\frac{n}{c}$ and $\forall x \in X$, $x$ is contained by at least one and to at most $k$ of them.(hence $r \le ks$) Let $K=\sum_{i=0}^{k} \binom{r}{i}$ and assume that $s>2k$. Prove that there exist two disjoint subsets $X_1$ and $X_2$ of $X$ s.t. $|X_i| \ge \frac{n}{2K}$ for both $i=1,2$, and none of the sets $A_1, A_2,...,A_r$ contains points from both sets $X_1$ and $X_2$.
Now I shall show you what I did: $\forall x \in X$, define the trace of it $T(x)=\left\lbrace i:x \in A_i \right\rbrace$. We partition the elements of $X$ by traces $X=\left\lbrace T_1 \cup T_2 \cup ... \cup T_{K} \right\rbrace$, the average size of these traces is $\frac{n}{K}$. Then at least $\frac{1}{2} |X|$ elements of $X$ belong to blocks(traces) of size at least $\frac{n}{2K}$. The last thing I need to do is to show that there exists two of these elements $x$ and $y$, s.t. $T(x) \cap T(y) = \emptyset$. Then I got stuck here.
I suppose that $\forall x, y$ of these elements, $T(x) \cap T(y) \ne \emptyset$ and try to find a contradiction with the size of each $A_i$, but I failed. Could you please give me some hints? Any help will be appreciated. Thanks a lot.