We flip a coin for each element of \(X\text{.}\) If the coin flip for \(x\) is heads, we colour it red and if it is tails we colour it blue. For a given set \(S\subseteq X\) of size \(n\text{,}\) the probability that every element of \(S\) is red is

\begin{equation*}
\left(\frac{1}{2}\right)^n\text{.}
\end{equation*}

Likewise, the probability that every element is blue is the same. Thus, the expected number of sets in \(\mathcal{F}\) whose elements all get the same colour is

\begin{equation*}
|\mathcal{F}|\left(\left(\frac{1}{2}\right)^n + \left(\frac{1}{2}\right)^n\right) = \frac{|\mathcal{F}|}{2^{n-1}}
\end{equation*}

which is less than one since \(|\mathcal{F}|\lt 2^{n-1}\text{.}\) Therefore, by the First Moment Principle, there must exist a colouring in which fewer than one set in \(\mathcal{F}\) is monochromatic. Since the number of monochromatic sets is an integer, it must be zero.