$$S = \{1, x , x^2\}$$
S is a set of orthogonal vectors. So instead of applying the gram-schmidt process to obtain the orthonormal basis, can't we just do
$$S' = \{1/||1||, x/||x||, x^2/||x^2||\}$$
Inner product is defined by
$$\int_{-1}^{1}p(x)q(x)dx$$
In your original problem, you are given a Euclidean vector space $\biggr(\mathbb{R}[x_2], (f, g):=\int\limits_{0}^{1}f(x)g(x)dx\biggr)$, where $\mathbb{R}[x_2]$ is a space of polynomials of degree 2 or less. Note that Euclidean vector space is already endowed with a special inner product $(f, g)$. You cannot define a Euclidean vector space without specifying a particular inner product that goes along with it. So in G-S you have to use this particular inner product, by the definition of Euclidean vector space and G-S process. Specifically, given basis $\{1, x, x^2\}$, let's find the orthogonalization $\{p_1, p_2, p_3\}$.
Apply G-S:
$p_1:=1$
$p_2:=x-\frac{(x, p_1)}{|p_1|^2}p_1=x-\frac{1}{2}$. Since $|p_1|^2=(p_1, p_1)=\int\limits_{0}^{1}1dx=1$ and $(x, p_1)=\int\limits_{0}^{1}xdx=\frac{1}{2}$.
Similarly,
$p_3:=x^2-\frac{(x^2, p_1)}{|p_1|^2}p_1 - \frac{(x^2, p_2)}{|p_2|^2}p_2 =x^2-x+\frac{1}{6}$.
Thus, the orthogonal basis is $\{1, x-\frac{1}{2}, x^2-x+\frac{1}{6}\}$.
Note that I did it for the range of integration from 0 to 1. You just need to compute the same objects for the range of -1 to 1.