This problem arised while I was working through Erdös-Bollobas's solution for Ramsey-Turán Problem however i believe that details for that isn't necessary.
Consider the following construction
\begin{equation} z, z^\prime, b, b^\prime \in S, \ \end{equation} and the $K_4$ formed by these points have the following property
\begin{equation} d(z, z^\prime) > 2 - t,\ d(b, b^\prime) > 2 - t, d(b, z^\prime) < \sqrt{2} - t, d(b^\prime, z^\prime) < \sqrt{2} - t, \ d(b^\prime, z) < \sqrt{2} - t, \ \end{equation}
I need to show that $d(b, z) \geq \sqrt{2} - t$. For some reason, I got stumped with this question, but it seems rather simple. Any help would be appreciated!
This follows from Ptolemy's inequality: $$ d(b,z) d(b',z') + d(b,z') d(b',z) \ge d(b,b') d(z,z'). $$ Suppose for the sake of contradiction that $d(b,z) < \sqrt2 - t$ as well. By replacing everything on the left by its upper bound, and everything on the right by its lower bound, we get $$ (\sqrt 2 - t)^2 + (\sqrt 2 - t)^2 > (2 - t)^2 $$ which (by solving a quadratic) is true when $t<0$ or when $t > 4\sqrt2-4$.
I assume that we are given $t \ge 0$, ruling out the first case. We also want $\sqrt2 - t$ to be positive, or else we're dealing with negative distances, which rules out the second case. So we cannot get $d(b,z) < \sqrt2 - t$.