For the supplied ordering $π$, let the cluster of $u$ be ${w | u ∈ wreach G π (2r) w}$, the set of vertices from which $u$ is weakly $2r$-reachable.

The degree bound is the definition read backwards: the set of clusters containing $v$ is indexed by ${u | u ∈ wreach G π (2r) v}$, which is $wreach G π (2r) v$ itself, of size at most $k$ by hypothesis. The radius bound drops the minimality clause: a vertex in the cluster of $u$ reaches $u$ by a walk of length at most $2r$, which reversed puts it in the $2r$-ball of $u$.

Covering is the only real argument. Given $v$, let $u$ be a $π$-minimal vertex of the $r$-ball of $v$, which is nonempty and finite. For $w$ in that ball, concatenate the reversed ball walk $w → v$ with the ball walk $v → u$: a walk of length at most $2r$ from $w$ to $u$. Every vertex on it lies on one of the two halves, and cutting a walk of length at most $r$ at any of its vertices shows that vertex to be within distance $r$ of both endpoints — so the whole support stays inside the $r$-ball of $v$, where $u$ is $π$-minimal. Hence $u$ is weakly $2r$-reachable from $w$, i.e. $w$ lies in the cluster of $u$.