Assume the contrary, for contradiction.
Let $A_1 \cup A_2 \cup \cdots \cup A_m$ be the largest group of mutually disjoint sets $A_1, A_2, \cdots, A_m \in F$, and suppose for contradiction that $m \le \frac{n-6}{3}.$ Let $a_1, a_2, \cdots, a_6$ be $6$ distinct elements which aren't in any of the $A_j$'s. Let $S_1, S_2, S_3$ denote $\{a_1, a_2\}, \{a_3, a_4\}, \{a_5, a_6\}$ respectively. Let $T_i$ be the set of elements $e \in A_1 \cup A_2 \cup \cdots \cup A_m$ for which $\{e\} \cup S_i \in F$, for each $i = 1, 2, 3.$ We know that $|T_i| \ge \left \lfloor \frac13 n \right \rfloor - 1$ by the condition and our assumption, since if $e \in T_i$ for $e \notin A_1 \cup A_2 \cup \cdots \cup A_m$, then we could add $\{e\} \cup S_i$ to the $A_i$'s to expand our group of mutually disjoint sets. Let $x_1, x_2, \cdots, x_m$ denote $|S_1 \cap A_1|, |S_1 \cap A_2|, \cdots, |S_1 \cap A_m|$ respectively. Define the $y_1, y_2, \cdots, y_m$ similarly for $S_2$, and the $z_1, z_2, \cdots, z_m$ for $S_3.$ By assumption, we have that $x_1 + x_2 + \cdots + x_m = |T_1| \ge \left \lfloor \frac13 n \right \rfloor - 1$, and similarly for $y_1 + y_2 + \cdots + y_m$, $z_1 + z_2 + \cdots + z_m.$ Therefore, we have that $(x_1+y_1+z_1) + (x_2+y_2+z_2) + \cdots + (x_m + y_m + z_m) \ge 3 (\left \lfloor \frac13 n \right \rfloor -1) \ge 3 (\frac{n-2}{3} - 1) = n-5.$ Therefore, since $m \le \frac{n-6}{3}$, we know by Pigeonhole that there is $1 \le k \le m$ with $x_k + y_k + z_k \ge 4.$
Suppose that $x_k \ge 2$ WLOG. Since $x_k \le 3$, we know that one of $y_k, z_k > 0$, suppose that $y_k \ge 1$ WLOG. Let $a, b \in A_k$ be so that $\{a\} \cup S_1, \{b\} \cup S_1$ are in $F$ and $c \in A_k$ be so that $\{c\} \cup S_2 \in F.$ Note that $c$ could be equal to one of $a$ or $b$. However, $c$ can't be equal to both $a$ and $b$, so suppose that $b \neq c.$ Then, if we replace $A_k$ with $\{b\} \cup S_1$ and $\{c\} \cup S_2$, we obtain a group of mutually disjoint sets $A_1, A_2, \cdots, A_{k-1}, A_{k+1}, \cdots, A_m, \{b\} \cup S_1, \{c\} \cup S_2$ which has strictly more sets, contradicting our initial assumption.
$\square$
