maximum size of intersecting family $F$ such that $|A∪B| < n$ for all $A,B ∈ F.$ Could someone please help me with this problem?
Suppose that $F \subset 2^{[n]}$ is an intersecting family such that $|A∪B| < n$ for all $A,B ∈ F.$ Prove that $|F| ≤ 2^{n−2}.$
I see that $|F| \leq 2^{n-1},$ since otherwise by pigeonhole principle we have a set $A$ and it's complement, which is a contradiction with the assumption of intersecting family. But I don't know how to use the assumption $|A∪B| < n$ to get desired bound.
Any help would be appreciated.
 A: We can prove it by contradiction, it follows directly from this lemma: (if there existed a family $F$ with more than $2^{n-2}$ elements then the set $F'$ of all sets that are a subset of at least one element of $F$ has at least $2^{n-1}+2$ elements and no two of its elements give all of $[n]$ as their union). Note that the proof of the linked result is a little bit similar to what I was trying to do below.

Flawed attempt ( in fact the lemma is false, consider taking $n=2k$, and letting $A$ just have the subset $\{1,2,\dots\,k\}$ and letting $B$ contain the $2^n-2^{k+1}+1$ suitable subsets).
Lemma: let $n\geq 2$ and let $A$ and $B$ be non-empty subsets of $2^{[n]}$ such that if $a\in A, b\in B$ then we have $a\cup b \neq [n]$ and $a\cap b\neq \varnothing$, then $|A|+|B|\leq 2^{n-1}$.
Proof: We proceed by induction, the base case is $n=2$, cleary $A$ and $B$ must both be equal and only contain one singleton. So we must have $|A|+|B|=2=2^{n-1}$.
Inductive step: Suppose it is true for $n$ and now let us prove it for $n+1$. Take $A$ and $B$ as in the theorem, now let $A_0=\{a | n\not\in a, a\in A\}$ and $A_1=\{a\setminus \{n\} | n\in a, a\in A\}$. Define $B_0$ and $B_1$ analogously. Notice that $A_0$ and $B_1$ satisfy the conditions of the theorem, because if we have $a\in A_0$ and $b\in B_1$ such that $a\cup b=[n-1]$ this would force the existance of two elements in $A,B$ such that their union is $[n]$, we also must have $a\cap b \neq \varnothing$. It follows that $|A_0|+|B_1|\leq 2^{n-1-1}$ by the induction hypothesis. Analogously $|A_1|+|B_0|\leq 2^{n-1-1}$. It follows that $|A|+|B|\leq 2^{n-1}$.
The gap is when one of the auxiliary sets is empty.
To conclude notice that what you want is just the particular result of this theorem in which $A=B$.
