Consider the following diagram, courtesy of mathinsight.org:
You can think of $(a \cdot u) u$ as the piece of $a$ that is in the direction of $u$. The part that is left over, $a - (a \cdot u) u$, must naturally be the missing side of the triangle, and hence is perpendicular to $u$. So at each step of the Gram-Schmidt process, the formula
$$ v_{n+1} = a - \sum_{j=1}^n \langle a, u_j \rangle u_j, \quad u_{n+1} = v_{n+1}/ \|v_{n+1} \|$$
does the following: it first subtracts all the pieces of $a$ that are in the same direction as all the $u_j$, then it renormalizes. The resulting vector must be orthogonal to all the $u_j$'s since you just subtracted out all the pieces that were not perpendicular.
Your $u_1$ is correct, but your $u_2$ is incorrect; as noted in the comments, $u_1\cdot u_2 \neq 0$.
Recall that the standard inner product on $\mathbb{C}^n$ is given by $\langle u, v\rangle = u\cdot\bar{v}$. With this in mind, let's calculate $u_2$. First we have
$$\langle v_2, u_1\rangle = (-1, i, 1)\cdot\overline{\frac{1}{\sqrt{2}}(1, 0, i)} = \frac{1}{\sqrt{2}}(-1, i, 1)\cdot(1, 0, -i) = \frac{-1-i}{\sqrt{2}},$$
so
$$w_2 = v_2 - \langle v_2, u_1\rangle u_1 = \left[\begin{array}{c} -1\\ i\\ 1\end{array}\right] + \frac{1+i}{\sqrt{2}}\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ i\end{array}\right] = \left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
As
\begin{align*}
\|w_2\|^2 &= \sqrt{\left|\frac{-1+i}{2}\right|^2 + |i|^2 + \left|\frac{1+i}{2}\right|^2}\\
&= \sqrt{\left(\frac{-1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0^2 + 1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2}\\
&= \sqrt{2}
\end{align*}
we have
$$u_2 = \frac{1}{\|w_2\|}w_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
Let's check to see if $u_2$ is orthogonal to $u_1$:
\begin{align*}
\langle u_1, u_2\rangle &= \frac{1}{\sqrt{2}}(1, 0, i)\cdot\overline{\frac{1}{\sqrt{2}}\left(\frac{-1+i}{2}, i, \frac{1+i}{2}\right)}\\
&= \frac{1}{2}(1, 0, i)\cdot\left(\frac{-1-i}{2}, -i, \frac{1-i}{2}\right)\\
&= \frac{1}{2}\left(\frac{-1-i}{2} + 0 + \frac{1+i}{2}\right)\\
&= 0.
\end{align*}
Note, we didn't have to normalise before we checked orthogonality; i.e. we could have checked $\langle u_1, w_2\rangle = 0$ instead.
I won't do the calculation of $u_3$ now. It is similar, but there are more computations. Note however that you have a typo in your formula for $w_3$; it should be
$$w_3 = v_3 - \langle v_3, u_1\rangle u_1 - \langle v_3, u_2\rangle u_2.$$
Now that you have the correct $u_2$ and the correct formula for $w_3$, the computation for $u_3$ should work out and produce $u_3 = \frac{1}{2}(i, -1-i, 1)$.
Best Answer
Start with $1$, and normalized it so that $\|\alpha 1\|=1$. So, $$ p_0 = \left(\frac{1}{\sqrt{2}}\right)1. $$ Next, take $x$ and subtract the projection of $x$ onto $p_0$: $$ x - (x,p_0)p_0 = x -\left(\frac{1}{2}\int_{-1}^{1}xdx\right)1 = x. $$ Then normalize to obtain $$ p_1 = \frac{1}{(x,x)^{1/2}}x=\frac{1}{\left(\frac{x^3}{3}|_{x=-1}^{x=1}\right)^{1/2}}x=\sqrt{\frac{3}{2}}x $$ Finally, start with $x^2$ and subtract the components along $p_0$ and $p_1$: $$ x^2 - (x^2,p_1)p_1 - (x^2,p_0)p_0 \\ = x^2 - \left(\int_{-1}^{1}t^2 \sqrt{\frac{3}{2}}tdt\right)\sqrt{\frac{3}{2}}x - \left(\int_{-1}^{1}t^2\frac{1}{\sqrt{2}}dt\right)\frac{1}{\sqrt{2}} \\ = x^2 - \left(\left.\frac{3}{2}\frac{t^4}{4}\right|_{t=-1}^{1}\right)x-\left(\left.\frac{1}{2}\frac{t^3}{3}\right|_{t=-1}^{1}\right)1 = x^2 - \frac{1}{3} $$ Then normalize \begin{align} p_2 & = \frac{1}{(x^2-1/3,x^2-1/3)^{1/2}}(x^2-1/3) \\ & = \frac{1}{\left(\int_{-1}^{1}(t^2-1/3)^2dt\right)^{1/2}}(x^2-1/3) \end{align}