Suppose $p_1p_2p_4p_6p_5p_3 p_1 \subset \mathbb{R}^2$ is a convex polygon (See figure ), let $d(p_1,p_2)=x$. Here $d(.,.)$ represents an Euclidean distance in $\mathbb{R}^2$. Given $$d(p_1,p_3) \geq x$$$$d(p_1,p_4) \geq x$$ $$d(p_2,p_3) \geq x$$ $$d(p_2,p_4) \geq x$$ $$d(p_3,p_4)< x $$
Now, I believe that if $d(p_5,p_6) \geq x$, then at least one of $d(p_5,p_3)$, $d(p_5,p_4)$, $d(p_6,p_3)$, $d(p_6,p_4)$ is greater than $x$. Otherwise, this polygon will lose its convexity either at $p_3$ or $p_4$. As illustrated in figure, polygon loses its convexity at $p_3$.
PS: The order of the points should be the same as the diagram as it makes more sense to the question.
Since $p_1p_2p_4p_6p_5p_3 p_1 \subset \mathbb{R}^2$ is convex, any ordered subsequence of $p_1p_2p_4p_6p_5p_3 p_1$ is convex also. In particular, $p_1p_2p_4 p_3 p_1$ is convex. Since $d(p_3, p_4) < d(p_1, p_2), $ the extension of the line $p_2 p_4$ meets the extension of the line $p_1 p_3$ at a point $Q$ which is closer to $p_3 p_4$ than $p_1 p_2.$ We are given that $d(p_6, p_5) \geq x.$ If $\min\{d(p_6, p_4), d(p_5, p_3)\}= d(p_6, p_4)\ $ as in the diagram below, then it is left to the reader to prove that $d(p_5, p_4) > d(p_5, p_6) = x.$ Similarly, if $\min\{d(p_6, p_4), d(p_5, p_3)\}= d(p_5, p_3)\ $ then a similar argument to the one just used to prove the previous sentence can be used to show that $d(p_6, p_3) > d(p_6, p_5) = x.$