[Solution] Since the shortest distance from a point to a line is the perpendicular segment from that point to the line, the minimum value of the required distance can be obtained by symmetry.
That is, choose the midpoints of $AB$ and $CD$ as $P$ and $Q$,
Since $PC$ and $PD$ are the altitudes of the equilateral triangles $ABC$ and $ABD$, then
$$
PC=PD=\frac{\sqrt{3}}{2} \text{. }
$$

Since $Q$ is the midpoint of side $DC$, we have $\quad QC=\frac{1}{2}$.
In the right triangle $\triangle CPQ$, by the Pythagorean theorem, we get
$$
PQ=\sqrt{PC^{2}-QC^{2}}=\sqrt{\frac{3}{4}-\frac{1}{4}}=\frac{\sqrt{2}}{2} .
$$

Therefore, the answer is $(C)$.
$23 \cdot 6$ In a regular tetrahedron with vertices $A, B, C, D$, each edge has length 1. Point $P$ is on edge $A B$, and point $Q$ is on edge $C D$. Find the shortest possible distance between points $P$ and $Q$.
(A) $\frac{1}{2}$.
(B) $\frac{3}{4}$.
(C) $\frac{\sqrt{2}}{2}$.
(D) $\frac{\sqrt{3}}{2}$.
(E) $\frac{\sqrt{3}}{3}$.
(30th American High School Mathematics Examination, 1979)