There is a standard proof of this written down someplace but I forget it.
You start out by finding the closest point Q on a line L1 to a specified point P.
You should notice that:
(I) Then line-segment QP is perpendicular to L1. Do you see why this must always be the case?
(II) If Q were not the closest point on L1 to P, then QP would not be perpendicular to L1.
Now consider that P is on another line L2 ...
1. If L2 is parallel to L1, then |PQ| is the distance of closest approach and PQ is perpendicular to both.
2. If L2 intersects L1, then |PQ| is not the distance of closest approach unless |PQ|=0.
3. If L2 is skew to L1 ... then we already have QP perpendicular to L1 beacuse Q is the closest point on L1 to P. But P is not the closest point to Q on L2 unless PQ is perpendicular to L2 ...
because of statement (II) above.
For |PQ| to be the distance of closest approach for the two skew lines, P must be the closest point on L2 to Q and Q must be the closest point on L1 to P ... so PQ must be perpendicular to both lines.
Since L1 and L2 are lines there can be only one (P,Q) pair such that |PQ| is the distance of closest approach - i.e. the solution is unique.