Let $\mathcal{F} = \{F_1,...,F_m \}$ be a family of subsets of $[n]$, such that for every $i \neq j, F_i \subsetneq F_j, F_i \cap F_j \neq \emptyset, \text{ and } F_i \cup F_j \neq [n]$. Prove that $m \leq {{n-1} \choose \lfloor \frac{n-1}{2} \rfloor}$.
Please comment on my attempt below.
Let $C_n$ denote the set of cyclic permutations on $[n]$, so $|C_n| = (n-1)!$. We derive the bound on $m$ by double-counting the set: $$M = \{(\pi,F): \pi \in C_n, F \in \mathcal{F} \text{ is an interval on } \pi \}$$
First, consider an arbitrary set $F \in \mathcal{F}$. The number of permutations containing $F$ as an interval is $|F|!(n-|F|)!$. This expression is smallest when $|F|$ deviates from $\frac{n}{2}$ as little as possible, i.e. $|F| = \lfloor \frac{n}{2} \rfloor =: k$. If we were to sum the number of permutations of $[n]$ this way over $F \in \mathcal{F}$, we'd get: $$|M| = \sum_{F \in \mathcal{F}}|F|!(n-|F|)! \geq |\mathcal{F}|k!(n-k)!$$
Now we count by fixing a permutation first. For an arbitrary $\pi$, the three conditions on $F_i$ and $F_j$ imply that the maximum number of sets $F$ that can occur as an interval on $\pi$ is $k$ (which happens when the size of the $F$'s is exactly $k$). Hence we have the bound: $$|M| \leq k|C_n| = k(n-1)!$$
Putting the two bounds together, we have: $$|\mathcal{F}|k!(n-k)! \leq k(n-1)!$$
And so: $$m = |\mathcal{F}| \leq \frac{(n-1)!}{(k-1)!(n-k)!} = {{n-1} \choose {k-1}} = {{n-1} \choose {\lfloor \frac{n}{2} \rfloor -1}} \leq {{n-1} \choose {\lfloor \frac{n-1}{2} \rfloor }}$$