Showing that an algorithm is 2-approximation
The place to start is at the simplest case: $k = 2$. Suppose that the points in $S$ are $x,y$. Consider some optimal solution of value $1$. We consider two cases.


$x,y$ belong to the same cluster. Thus $d(x,y) \leq 1$. Since $y$ is the farthest point from $x$, all other points are at distance at most $1$ from $x$, and in particular at distance at most $1$ from $\{x,y\}$. This means that in the clusters generated by the algorithm, each point is at distance at most $1$ from its base (either $x$ or $y$), and so the diameter of each cluster is at most $2$.
$x,y$ belong to different clusters. In this case, since the optimal solution has value $1$, each point other than $x,y$ is within distance $1$ of either $x$ or $y$ (the one which shares cluster with it). This means that in the clusters generated by the algorithm, each point is at distance at most $1$ from its base, and so again the diameter of each cluster is at most $2$.


See if you can generalize these ideas for $k &gt; 2$.