Suppose $j=3$ then $a_3,b_3$ are two distinct numbers.
$a_1$ and $a_2$ are both less that $a_3$ so $a_3$ must be at least $3$
$b_4$ and $b_5$ are both less than $b_3$ so $b_3$ must be at least $3$
but we can't have $a_3=b_3$ so one of them must be at least $4$. That's what we get from the first attempt. But we can do better.
Now suppose $a_3$ and $b_3$ are both less than $6$ then $a_1, a_2, b_4, b_5$ (the four numbers referred to in the proof you have quoted) must all be less than $6$ too - that is six distinct positive integers less than $6$, which is impossible, so the largest, which will be $a_3$ or $b_3$, will be at least $6$.
The proof you have been given takes this observation and generalises it to any $j$ rather than just $3$. Always try to work out a specific example of a general statement if you don't quite understand what it is saying.
Another way of approaching this is to look at the way the numbers $6,7,8,9,10$ are distributed between the $a_i$ and $b_i$. If three are $b$s then two must be $a$s, for example.