Hello, and thanks for this informative answer. I recently ran into this question as well and read your answer. But I'm not familiar with Berger's lemma about geodesics realizing the diameter, and when I searched it, I couldn't find it. Is it possible for you to either state the lemma or give references to this? I've John Lee's two books on Riemannian geometry, Do Carmo, Sakai and your book on metric geometry.

@MoisheKohan Thank you for your comment! Yes indeed, convex means indeed path connected, so connected, so that was redundant. Modified the question now! Also, an axis of symmetry is a line $l$ in the ambient Euclidean space with respect to which the reflection $R_l$ leaves the space invariant, i.e. if $x\in K \iff R_l(x) \in K.$ Added this bit into the question for further clarification.

@YCor You're right! Let me think how to rephrase the question!

@YCor Apologies for any possible misunderstanding of your comment, but I read your statement as no need to refer to $K$ as convex. But I needed the convexity of $K$ to guarantee $T_0K$ is convex. I think you were asking me to rephrase the question for generalized convex cones, instead of tangent cones to convex open subsets. Is this correct?

@YCor: if $K$ is just the union of nonnegative x and y-axes in $\mathbb{R}^2,$ is its tangent cone at the origin, $T_0K$ convex?

@Raziel Thanks for these papers (not my question). I just looked at the first paper, but there they assumed that $$ is a submanifold of the Euclidean space, not a general manifold. Will the argume,nts there generalize from Euclidean space to general $M?$

I wonder we can say something more about the differentiability of $p\mapsto d(p,S)$ aside from almost everywhere differentiable? I feel that it can be smooth aside from a set of measure zero. Also, aside from the subset $S$ itself, are there any other point $p\mapsto d(p,S)$ is not differentiable?