I seem to be getting different results depending on whether I use OP's suggested inner product in the comments, or if I use the polarization identity.
Before everything else, however, here is a routine by Velvel Kahan for evaluating the divided difference of a polynomial, based on Horner's method:
polynomialDividedDifference[poly_, {x_, a_, b_}] /; PolynomialQ[poly, x] :=
Module[{d = 0, y = 0},
Do[y = b y + Coefficient[poly, x, k]; d = a d + y,
{k, Max[0, Exponent[poly, x]], 1, -1}];
Expand[d]]
With this, here is the result of using the inner product
$$\langle f,g\rangle=\int_{-1}^1\int_{-1}^1\frac{(f(u)-f(v))(g(u)-g(v))}{(u-v)^2}\,\mathrm du\mathrm dv$$
Orthogonalize[x^Range[4],
Integrate[polynomialDividedDifference[#1, {x, u, v}]
polynomialDividedDifference[#2, {x, u, v}],
{u, -1, 1}, {v, -1, 1}] &]
{x/2, 1/2 Sqrt[3/2] x^2, 3/2 Sqrt[5/13] (-((2 x)/3) + x^3),
15/8 Sqrt[21/31] (-((14 x^2)/15) + x^4)}
(same as in Michael's answer)
Contrast this with the result of using the inner product
$$\langle f,g\rangle=\frac14\left(\left(\int_{-1}^1\int_{-1}^1\frac{(f(u)-f(v)+g(u)-g(v))^2}{(u-v)^2}\,\mathrm du\mathrm dv\right)^2-\left(\int_{-1}^1\int_{-1}^1\frac{(f(u)-f(v)-(g(u)-g(v)))^2}{(u-v)^2}\,\mathrm du\mathrm dv\right)^2\right)$$
based on the polarization identity:
Orthogonalize[x^Range[4],
(Integrate[polynomialDividedDifference[#1 + #2, {x, u, v}]^2,
{u, -1, 1}, {v, -1, 1}]^2 -
Integrate[polynomialDividedDifference[#1 - #2, {x, u, v}]^2,
{u, -1, 1}, {v, -1, 1}]^2)/4 &]
{x/8, (3 x^2)/16, (1036800 (-((719 x)/2880) + x^3))/3838561,
(17920000 (-((3177 x^2)/6400) + x^4))/33294461}
The OP will have to choose which basis is most appropriate for his situation.