Consider the function, prove that it's bilinear, symmetric, and positive definite

$F$ is clearly symmetric, as \[ F(p(x),q(x)) = \int_{-1}^1 p(x)q(x)dx = \int_{-1}^1 q(x)p(x)dx = F(q(x),p(x)). \]  Given that, it's enough to check that it's linear in one of the two entries to show that's bilinear. This is clear, as \[ F(ap(x) + bq(x), r(x) = \int_{-1}^1 (ap(x)+bq(x))r(x) dx = a \int_{-1}^1 p(x)r(x)dx + b \int_{-1}^1 q(x)r(x)dx = aF(p(x),r(x)) + bF(q(x),r(x)). \] Finally, for every polynomial $p(x)$, \[ F(p(x), p(x)) = \int_{-1}^1 p(x)^2 dx \geq 0 \] (as $p(x)^2 \geq 0$) so it's also positive definite.

It is clear that $p_0$, $p_1$ and $p_2$ form a basis of the polynomials of degree $\leq 2$ (they're independent and it's a $3$-dimensional space). To check that the basis is orthogonal, we compute \[ F(p_0(x), p_1(x)) \int_{-1}^1 p_0(x)p_1(x)dx = \int_{-1}^1 x dx = 0 \] because $x$ is an odd function over a symmetric interval; similarly, \[ F(p_1(x), p_2(x)) \int_{-1}^1 p_1(x)p_2(x)dx = \int_{-1}^1 \left( \frac{3}{2}x^3 - \frac{1}{2}x \right) dx = 0 \] as it is also an odd function over a symmetric interval. Finally, \[ F(p_0(x), p_2(x)) \int_{-1}^1 p_0(x)p_2(x)dx = \int_{-1}^1 \left( \frac{3}{2}x^2 - \frac{1}{2} \right) dx = \left[ \frac{1}{2} x^3 - \frac{1}{2} x \right]_{-1}^1 = 0 \] as the function equals $0$ when $x=1$ and $x=-1$.