Here's the question: Let S be a set which $S=\{1,2,\ldots,n\}$. Let $A_1,A_2,\ldots,A_n$ be subsets of $S$ which aren't equal. We need to prove there is an $x$ in $S$ such that the union of $x$ with $A_1,A_2,$ $\ldots,A_n$ keeps them not equal. There's a clue: To build a graph that it's vertices are $A_1,A_2,\ldots,A_n$. Now I've thought what the edges should be, and couldn't find the correct condition.
Thanks, Sergey
If it is not necessary to prove with graphs I can suggest you another proof. Let's $$|A_1|\le|A_2|\le\cdots\le|A_n|.$$ Suppose it is not true, so for every $k\in\{1,2,...,n\}$ there are $i_k,j_k\in\{1,2,...,n\}$, such that $A_{i_k}=A_{j_k}\setminus\{k\}$,i.e. $|A_{i_k}|=|A_{j_k}| - 1$. It is easy to see that it is contradiction.