SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $A_1,\dots ,A_6$ be the subsets. We induct on the number $n$ of elements of $X$.

If $n=5$, since $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10>6$, we can find a three-element subset $Y$ of $X$ not equal to any of $A_1,\dots ,A_6$; coloring the elements of $Y$ in one color and the other elements in the other color.

If $n=6$, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20>6\times 2=12$, we can find a three-element subset $Y$ of $X$ not equal to any of $A_1,\dots ,A_6$ or its complements; coloring the elements of $Y$ in one color and the other elements in the other color meets the desired condition.

Now suppose $n\ge 7$. There must be two elements $u,v$ of $X$ such that $\{u,v\}$ is not a subset of any $A_i$, since there are at least $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ pairs, and at most $$6\times \left|\{(x,y)\mid x,y\in A_i,i=1,\dots ,6\}\right|=6\times 3=18$$ lie in an $A_i$. Replace all occurrences of $u$ and $v$ by a new element $w$, and color the resulting elements using the induction hypothesis.

Now color the original set by giving $u$ and $v$ the same color given to $w$.

$\square$