Suppose we have a 'difference function' D on the space of colours, a simple metric d on the target space, and some nonlinear function f from the space of colours to the target space such that D(x,y) = d(f(x),f(y)) for all x and y.

Then the triangle inequality for D would be D(x,y) + D(y,z) ≥ D(x,z). By the equation above this is equivalent to d(f(x),f(y)) + d(f(y),f(z)) ≥ d(f(x),f(z)), which is true by the triangle inequality in the target space at the points f(x), f(y) and f(z). If the target space uses a simple metric like the Euclidean or Manhattan distance then the triangle inequality is always going to hold in the target space and hence in colour space.

Note that the above argument does not need linearity of f.

The sphere example doesn't hold up because if two points are further apart than half way around a great circle then the distance between them is actually the length of the path going the other way.