Maximum distance between points in a triangle

Re your PS: Pick two points $P,Q$ in an equilateral triangle $A,B,C$. project $A,B,C$ to the line $PQ$, giving $A',B',C'$. Say $A'$ is the "leftmost" and $B'$ the "rightmost" of the projections. Then $|PQ|\le |A'B'|\le |AB|$. (We see that by the same argument, two points in a polygon are at most as far apart as two vertices, i.e., as the maximal edge-or.diagonal.

Now to c: Let $ABC$ be our triangle. Pick suitable $r=\frac{\sqrt3-1}{2}\approx 0.366<\frac12$. Draw the circles with radius $r$ around $A,B,C$. In the "middle" there is room for a small equilateral triangle with vertices on these circles and side length $1-r\sqrt 3=r$. Verify that all seven parts in the figure have diameter $r$. Therefore, the pigeons strike back if we have eight points.

To visualize, in the following illustration all blue circles have radius $r$ and centre one of the nodes: