Prove that there is $i$ for which $i\in B$ and $i+1\in A$ Let $n \geq 3, \ A,B \subseteq \{2,...,n-1\}, \ A,B \neq \emptyset $ and $|A|+|B| \geq n$. Show that there is such $i$ that $i\in B $ and $i+1 \in A$.
My attempt:
Since $\ A,B \subseteq \{2,...,n-1\}$ and $|\{2,...,n-1\}| = n-2$ we have $|A \cup B| \leq n -2$ (because there is at most $n-2$ distinct elements).
Let $X=\{a-1: a\in A\}$. Then $X \subseteq \{1,...,n-2\}$. Of course, $|X|=|A|$. Since we added one new element which can be in $X \cup B$ (and from the fact that $B \subseteq \{2,...,n-1\}, \ X \subseteq \{1,...,n-2\}$), we have $|X \cup B| \leq n-1$ since in this sum of sets there can be at most $n-1$ distinct values.
Now
$$|X \cap B| = |X| + |B| - |X \cup B|$$
$$|X \cap B| = |A| + |B| - |X \cup B|$$
Since
$$ |X \cup B| \leq n -1$$
$$ -|X \cup B| \geq -n +1$$
$$ n-|X \cup B| \geq n-n +1$$
$$ n-|X \cup B| \geq 1$$
Because $|A| + |B| \geq n$ we can write
$$ |X \cap B| =|A| + |B|- |X \cup B|  \geq n-|X \cup B| \geq 1$$
So
$$ |X \cap B| \geq 1$$
So there is $i \in X \cap B$. Finally, $i\in X \implies i+1 \in A$.
 A: I dispute the statement that:
because $|A| + |B| \geq n,$
that you must then have that 
(in effect) $(A \cup B) = \{2,3,\cdots, (n-1)\}.$
That is, I see no reason that $A$ and $B$ can't have a sufficient number of shared elements, that then $|A| + |B| \geq n$ is satisfied, even though $(A \cup B)$ is not equal to $\{2,3,\cdots, (n-1)\}.$

My alternate proof is the following:
Assume that $|B| = r \implies r \in \{1,2,\cdots, (n-2)\}.$
Order the distinct elements in $B$ in ascending order, 
as $a_1, a_2, \cdots, a_r$
so that $a_1 < a_2 < \cdots < a_r.$
For $i \in \{1,2,\cdots, r\}$ 
let $b_i$ denote $(a_i + 1).$
Assume that the constraint that there exists an element $x$ such that $x \in B$ and $(x+1)\in A$ is violated.
Edit
Without loss of generality, $r > 1$, else the max value of $|A| + |B| = (n-2) + 1 < n$.
Then, none of $b_1, b_2, \cdots, b_{r-1}$ can be in $A$, and each of $b_1, b_2, \cdots, b_{r-1}$ must represent a distinct element in $\{2,3,\cdots, (n-1)\}.$
Therefore, the most elements that the set $A$ can have is 
$[(n-2) - (r-1)]$.
This implies that the largest value achievable by 
$|A| + |B|$
is $r + [(n-2) - (r-1)] = (n-1).$
