I'll give two proofs.

Divide the circle into six equal circular sectors. If necessary, rotate the sectors around the center until one of the points (say, A) falls on one of the radii. Note that, since each sector is a portion of the Reuleaux triangle of width 1, no points in a sector are father from each other than 1.

If in one of the two sectors to which A belongs there is another point of the set, these two will be within 1 of each other. Otherwise, we are left with 4 sectors and 5 points and, thus, are able to claim that one of the sectors contains at least two points of the set. These two then satisfy the requirements.

The second proof is by William A McWorter Jr..

We may assume that no point is at the center O of the circle and that no two of the points are on a radius; for then we are done. Radii through O and each of the six points partition 360^o into six parts, at least one of which is less or equal 60^o. Let A and B be two of the points such that the AOB is less or equal 60^o. AO and BO, being on radii, have length less or equal 1. But also, one of these segments is opposite an angle greater or equal 60^o in the AOB. Hence the length of AB is less or equal that of one of AO or BO, both of which are less or equal 1.