It is well known that if a weight function of a graph satisfies the triangle inequality, then 2 vertex connectivity is same as 2 edge connectivity in that graph. Can someone find a proof for this?
I looked at many papers, but couldn't find anything clear about the proof.
Connectivity and triangle inequality for distance function don't have anything common. There is a graph $G = \overline{\overline{2K_2}\cup K_1}$ that is 2-edge-conncted but is not 2-vertex-connected. You may use any distance function for it.
So, $G = (\{\,1, 2, 3, 4, 5\,\}, \{\,\{\,1, 2\,\}, \{\,3, 4\,\}, \{\,1, 5\,\}, \{\,2, 5\,\}, \{\,3, 5\,\}, \{\,4, 5\,\}\,\})$. Let $d(u, v) = A_{uv}$ for $$A = \begin{bmatrix} 0 & 1 & 2 & 2 & 1\\ 1 & 0 & 2 & 2 & 1\\ 2 & 2 & 0 & 1 & 1\\ 2 & 2 & 1 & 0 & 1\\ 1 & 1 & 1 & 1 & 0 \end{bmatrix}.$$
Probably it was meant that 2-vertex-connected graph is 2-edge-connected. That is true since $\kappa(G) \le \lambda(G)$ for any graph $G$, and this faact has no relation to distance function.