In this question, it is asked whether for any triangulation $C$ of a sphere $S^k$, and for any vertex $v$ of $C$, the link of $v$ is homeomorphic to a sphere $S^{k-1}$. This answer shows a concrete example of a triangulation of $S^5$ that does not have this property.
MY QUESTION: given a length parameter $\delta$, does there always exist some triangulation of $S^k$ with simplices of diameter at most $\delta$, that has the above property?
My guess is that the barycentric triangulation should satisfy this property, but I do not know how to prove it (if it is true).