Let $0 < p < q < 1$. Notice that $p^* = \frac{q-p}{1-p} \in (0, 1)$. We construct the sets $X$ and $Y$ as follows: For each element $i \in \{1, 2, \dots, n\}$ we put $i$ in $X$ with probability $p$ (independently of each other), and for each element $j \in \{1, 2, \dots, n\} \setminus X$ we put $j$ in $Y$ with probability $p^*$. Then $P(x \in X \cup Y) = p + (1-p)p^* = q$. It is easy to see that $f(p) = P(X \in \mathcal{A})$, and $f(q) = P(X \cup Y \in \mathcal{A})$, but since $\mathcal{A}$ has the property from the condition we have that $X \subseteq X \cup Y \Rightarrow f(p) < f(q)$.
$\square$