Note that $BP=BQ=DR=DS=1-s.$ It follows that $\triangle BPQ$ and $\triangle DRS$ are congruent isosceles right triangles.
In $\triangle BPQ,$ we have $PQ=BP\sqrt2,$ or 
\begin{align*} s &= (1-s)\sqrt2 \\ s &= \sqrt2 - s\sqrt2 \\ \left(\sqrt2+1\right)s &= \sqrt2 \\ s &= \frac{\sqrt2}{\sqrt2 + 1}. \end{align*}
Therefore, the answer is \[s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.\]
~MRENTHUSIASM
Square $ABCD$ has side length $1$. Points $P$, $Q$, $R$, and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s$. What is $s$?
$\textbf{(A) } \frac{\sqrt{2}}{3} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } 2 - \sqrt{2} \qquad \textbf{(D) } 1 - \frac{\sqrt{2}}{4} \qquad \textbf{(E) } \frac{2}{3}$