We can imagine some clustering problem with $k$-centers. For $G = (V,E)$ with distance $d_{ij} \ge 0$ between each pair of vertices $i,j \in V$. We assume that $d_{ii} = 0, d_{ij} = d_{ji}, d_{ik} + d_{kj} \ge d_{ij}$ for each $i,j,k \in V$. Now, problem is find $S \subset V$ which $|S| = k$. Which makes $max(d(j,S))$ smallest. It’s like making clusters efficiently.
There is a $2$-approximation algorithm for this. Let’s define $d(i,S) = \min_{j \in S} d_{ij}$.
$S \leftarrow$ {$i$}
$\operatorname{while}$ $|S| < k$ $\operatorname{do}$
$S \leftarrow S \cup$ {$j$}
It’s running time is trivial to be in polynomial time. Now, let’s define that $O$ $=$ {$e_1, e_2, \cdots, e_k$} is an optimal solution, $S^O$ $=$ {$S_1, S_2, \cdots, S_k$} as each cluster and $r^{\star}$ as its radius. Then two points in each $S_i$ can’t be greater than $2r^{\star}$ by triangle inequality.
Now, let’s think about the soltuion of an approximation algorithm $A$. Between the progress, if $A \neq O$ then in some moment algorithm needs to choose two points from some $S_i$. However, it means that farest points are already bounded in $2r^{\star}$ from the fact above. As a result, it can’t make bigger radius than $2r^{\star}$.