k-center problem: proof for Gon algorithm gives a 2-approximation
If $k+1$ vertices are at least $h$ units apart, then no ball of diameter $&lt; h$ can cover two of these vertices, hence you would need at least $k+1$ balls.
Since you want to have at most $k$ balls, $h$ must be at least the minimum distance of the $k+1$ vertices (pairwise).
Another way of phrasing it would be:

If you have a set $T$ of $k+1$ points, and at most $k$ balls of diameter $h$, then at least one ball must contain two points.  Let $d$ be the shortest distance between two vertices in $T$ (the two closest vertices).  Then $h \geq d$.


Taking the contra-positive of the lemma:  If $\text{OPT} &lt; h$, then you cannot have $k+1$ points that have pairwise distance $\geq h$, or in other words, if  $\text{OPT} &lt; h$ then

Any set with pairwise distance at least $h$ has at most $k$ vertices
Any set with at least $k+1$ vertices has a pair with distance less than $h$.