You're looking to prove that if $q$ violates condition (2) for $[a_1,a_3,\dots,a_j]$ then it also violates (2) for $[a_1,a_2,\dots,a_j]$, which reduces to showing that $d(q,a_1)\leq r$. Your hint, then, is to convince yourself that the "b-condition" is a sufficient condition to show that $d(q,a_2) = d_\mathbb{R}(q,a_2)$. Informally: there are no shortcuts through $p$ that let you route around $a_1$.

You're looking to prove that if $q$ violates condition (2) for $[a_1,a_3,\dots,a_j]$ then it also violates (2) for $[a_1,a_2,\dots,a_j]$, which reduces to showing that $d(q,a_2)\leq 2r$.

If $q$ is not the distinguished point $p$, then $2r\geq d(q,a_3) = d(q,a_2)+d(a_2,a_3)$, with the last equality following because the metric space is almost linear.

So the case you need to check more carefully is $q=p$, the distinguished point.