Answer: $1-\frac{1}{\sqrt{3}}$ Clearly there must be two sides with two points of the hexagon and two sides with one point of the hexagon. If the two sides with two points are adjacent then the sum of two of the angles would be $270^{\circ}$. Thus WLOG assume that the square is $ABCD$ the hexagon is $P_1P_2P_3P_4P_5P_6$ with $P_1,P_2\in AB$, $P_3\in BC$, $P_4,P_5\in CD$, and $P_6\in DA$. Then $$P_1P_2\;+\;P_4P_5=2\;-\;(AP_1\;+\;BP_2\;+\;CP_4\;+\;DP_5)=2\;-\;\frac{1}{\sqrt{3}}(AP_6\;+\;BP_3\;+\;CP_3\;+\;DP_6)=2(1\;-\;\frac{1}{\sqrt{3}})$$Thus the minimum side is at most $1-\frac{1}{\sqrt{3}}\;$ this is achieved when $P_3$ and $P_6$ are the midpoints of $BC$ and $DA$.