Prove as shown in the figure, connect the midpoints $H$ and $G$ of $AB$ and $CD$. On the line segment $GH$, take $PH = \frac{\sqrt{3}}{2}$. It is easy to see that
$$\begin{aligned} PA & = PB = 1, \\ PC & = PD = \sqrt{\left(\frac{1}{2}\right)^{2} + \left(1 - \frac{\sqrt{3}}{2}\right)^{2}} \\ & = \frac{1}{2}(2 - \sqrt{3}). \end{aligned}$$

Thus, point $P \in$ the intersection of the four circles $S$.
Similarly, on $GH$, take $GQ = \frac{\sqrt{3}}{2}$, then $Q \in S$.

And
$$\begin{array}{l} PQ = PH - QH \\ = \frac{\sqrt{3}}{2} - \frac{1}{2}(2 - \sqrt{3})^{4} = \sqrt{\frac{H}{3} - 1} \end{array}$$

satisfies the given condition.
Prove on the line connecting the centers of the top and bottom faces of the cube $\mathrm{OO}_{1}$, take point $P$ such that $PO = \frac{\sqrt{2}}{2}$. It is easy to see that $PA = PB = PC = PD = 1$,
$$PA_{1} = PB_{1} = PC_{1} = PD_{1} = 2 - \sqrt{2} < 1,$$

i.e., point $P \in$ the intersection $S$.
Similarly, take $QO_{1} = \frac{\sqrt{2}}{2}$, $Q \in OO_{1}$, then $Q \in S$.
$$\text{Thus, } \begin{aligned} PQ & = PO - QC. \\ & = \frac{\sqrt{2}}{2} - \frac{1 - \sqrt{2}}{2} = \sqrt{2} - 1 \end{aligned}$$

satisfies the given condition.