smc

Note that the distance $PQ$ will be minimized when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find this distance, consider triangle $△PCQ$. $Q$ is the midpoint of $CD$, so $CQ=\frac{1}{2}$. Additionally, since $CP$ is the altitude of equilateral $△ABC$, $CP=\frac{\sqrt{3}}{2}$. Next, we need to find $\cos (\angle PCQ)$ in order to find $PQ$ by the Law of Cosines. To do so, drop down $D$ onto $△ABC$ to get the point $D^{\prime }$. $\angle PCD$ is congruent to $\angle D^{\prime }CD$, since $P$, $D^{\prime }$, and $C$ are collinear. Therefore, we can just find $\cos (\angle D^{\prime }CD)$. Note that $△C{D}^{\prime }D$ is a right triangle with $\angle C{D}^{\prime }D$ as a right angle. As given by the problem, $CD=1$. Note that $D^{\prime }$ is the centroid of equilateral $△ABC$. Additionally, since $△ABC$ is equilateral, $D^{\prime }$ is also the orthocenter. Due to this, the distance from $C$ to $D^{\prime }$ is $\frac{2}{3}$ of the altitude of $△ABC$. Therefore, $C{D}^{\prime }=\frac{\sqrt{3}}{3}$. Since $\cos (\angle D^{\prime }CD)=\cos (\angle PCQ)=\frac{C{D}^{\prime }}{CD}$, $\cos (\angle PCQ)=\frac{\frac{\sqrt{3}}{3}}{1}=\frac{\sqrt{3}}{3}$ $$P{Q}^2=C{P}^2+C{Q}^2-2(CP)(CQ)\cos (\angle PCQ)$$ $$P{Q}^2=\frac{3}{4}+\frac{1}{4}-2\left(\frac{\sqrt{3}}{4}\right)\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{3}\right)$$ Simplifying, $P{Q}^2=\frac{1}{2}$. Therefore, $PQ=\frac{\sqrt{2}}{2}\Rightarrow$ $\boxed{\frac{\sqrt{2}}{2}}$