Since official solution is easy to find,I will post an alternative approach:
Proof:Of course,we use induction.Base case is trivial
Denote the subsets of $ U$ are $ A_1,A_2,...A_{n-2}$
Firstly,We can simply assume $ |A_i|=2$ for all $ i,1\le i \le n-2$
Because if $ |A_i|>2$ we can choose a $ 2-element$ subset $ A_i'$ of $ A_i$,and replace $ A_i$ by $ A_i'$
Secondly if every element of $ U$ appears in $ A_1,A_2,...A_{n-2}$ at least twice
Then $ 2n-4=|A_1|+...+|A_{n-2}|\ge 2n$ A contradiction!
This implies one element(WLOG,let it be $ 1$) appears in $ A_1,A_2,...A_{n-2}$ exactly once or never
So we divided the problem into two cases
Case 1:$ 1$ doesn't appear in $ A_1,A_2,...A_{n-2}$
delete $ 1$ from $ U$ and applies induction hypothesis to $ U'$ and $ A_2,...,A_{n-2}$
We got an arrangement of $ 2,3,...,n$ splits $ A_2,...A_{n-2}$ and add $ 1$ between the elements of $ A_1$
get our result!
Case 2:$ 1$ appears in $ A_1,A_2,...A_{n-2}$ exaxtly once (WLOG,let $ 1$ appear in $ A_1=\{1,a\}$
delete $ 1$ from $ U$ and applies induction hypothesis to $ U'$ and $ A_2,...,A_{n-2}$
We got an arrangement of $ 2,3,...,n$ splits $ A_2,...A_{n-2}$ and add $ 1$ far away from $ a$
get our result!
QED