Instead of counting shared corner atoms, pick one atom sitting exactly at a face centre of the FCC cell and count how many other atoms sit the same closest distance away from it. In an FCC lattice, atoms actually touch each other along a face diagonal, so travelling along each of the face-diagonal directions that pass through the chosen atom, in the three mutually perpendicular planes meeting at that atom, there is exactly one nearest neighbour per direction. Each of these three planes contributes four such touching neighbours around the atom, so \( 4 \times 3 = 12 \) atoms are found to be in direct contact with it. This matches the densest possible packing of equal spheres, which is exactly why both FCC and hexagonal close packed lattices share the same coordination number of 12. So the correct choice is option (D).