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

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

If $n = 6$, since $\binom{6}{3} = 20 > 6 \times 2 = 12$, we can find a three-element subset $Y$ of $X$ not equal to any of $A_1, \ldots, A_6$ 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 $n \geq 7$. 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 $\binom{7}{2} = 21$ pairs, and at most
$$6 \times \left| \{ (x, y) \mid x, y \in A_i, i = 1, \ldots, 6 \} \right| = 6 \times 3 = 18$$
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 $w$.

$\square$