Is there always a shortest path using vertices of low eccentricity? We say a graph $G$ is self-centered if $\text{rad}(G)=\text{diam}(G)$, so if its radius equals its diameter. In other words, the eccentricity of every vertex is equal.
Consider the following claim: let $G$ be a graph that is not self-centered, and let $u$ and $v$ be two distinct arbitrary vertices both with eccentricity equal to $\text{diam}(G)$. Then, there exists a shortest $u$-$v$ path on which $u$ and $v$ are the only vertices with maximum eccentricity, i.e. $\epsilon(u) = \epsilon(v) = \text{diam}(G)$.
If $u$ and $v$ are adjacent, this is trivially true. I'm not sure why this would always necessarily hold. Is there a proof or a counterexample?

This is not true.
Let $G$ be a wheel graph with at least 8 spokes and subdivide each spoke.
You can easily see that the center vertex is the only central vertex of $G$
and the vertices on the rim are the only vertices with maximum eccentricity (4).
However, if you take two vertices on the rim that are at distance two, then the
only shortest path between them goes along the rim and consists exclusively of vertices with maximum eccentricity.
