The missing clue you ask for might be that:
For every random variable $Z$ and positive real number $z$, if $P(|Z|\ge z)\le\frac13$ then $|m(Z)|\le z$.
This follows directly from the definition of the median. In the context of your question, using the inclusion $ [\,|Y_j-Y_n|\ge2ub_n\,]\subseteq [\,|Y_j|\ge ub_n\,]\cup[\,|Y_n|\ge ub_n\,], $ which implies the inequality $ P(|Y_j-Y_n|\ge2ub_n)\le P(|Y_j|\ge ub_n)+P(|Y_n|\ge ub_n), $ a consequence of the missing clue is that it is enough to prove that:
(0) For every positive real number $u$, there exists a finite integer $N$ such that for every $n\ge N$ and every $j\le n$, $P(|Y_j|\ge ub_n)\le\frac16$ and $P(|Y_n|\ge ub_n)\le\frac16$.
We now prove (0). Fix a positive real number $u$.
First, when $n\to+\infty$, $P(|Y_n|\ge ub_n)\to0$, hence there exists a finite $K$ such that:
$\qquad$(i) for every $n\ge K$, $P(|Y_n|\ge ub_n)\le\frac16$.
Second, for each $j\le K$, $P(|Y_j|\ge ub_n)\to0$ when $n\to+\infty$ because $b_n\to\infty$, hence there exists a finite $M_j$ such that:
$\qquad$(ii) for every $n\ge M_j$, $P(|Y_j|\ge ub_n)\le\frac16$.
Choose finally $N\ge K$ such that $N\ge M_j$ for every $j\le K$. Assume that $n\ge N$ and that $j\le n$.
- Using (i), one sees that $P(|Y_n|\ge ub_n)\le\frac16$ because $n\ge K$.
- If $j\le K$, using (ii), one sees that $P(|Y_j|\ge ub_n)\le\frac16$ because $n\ge M_j$.
- If $j\ge K$, using (i) once again, one sees that $P(|Y_j|\ge ub_n)\le P(|Y_j|\ge ub_j)\le\frac16$ because $b_j\le b_n$.
Hence the proof of (0) is complete.