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 $u_3 = \frac{1}{2}(i, -1-i, 1)$.
Orthonormal bases are nice because several formulas are much simpler when vectors are given wrt an ON basis.
Example: Let $\mathcal E = \{e_1, \dots, e_n\}$ be an ON basis. Then the Fourier expansion of any vector $v\in\operatorname{span}(\mathcal E)$ is just $$v = (v\cdot e_1)e_1 + (v\cdot e_2)e_2 + \cdots + (v\cdot e_n)e_n$$
Notice that there are no normalization factors and we don't need to construct a dual basis -- it's just a really simple formula.
In your example, of course $\{(1,0),(0,1)\}$ spans the same space as $\{(3,2),(2,2)\}$. But let me provide an example of my own: what about $\{(1.1,1.2,0.9,2.1,4),(3,-2,6,14,2),(6,6,6,3.4,11.1)\}$? There's certainly no subset of the standard basis vectors that spans the same space as these linearly independent vectors. But this is a pretty poor choice of basis because they're not orthonormal. It'd sure be nice if we had some algorithm that could produce as ON basis from them...
Best Answer
Your original vector $w_3$ is a linear combination of the previous two; in fact, $w_3 = w_1 + w_2$. Whenever that happens, the Gram-Schmidt process will spit out the zero vector. (Because $v_3$ will be forced to be in the span of $w_1$ and $w_2$, but also orthogonal to $w_1$ and $w_2$, the only possibility for $v_3$ is $0$.)
Go back and produce a basis for your subspace, then apply the Gram-Schmidt process and you'll have an orthogonal basis as desired.