Take $T=\begin{bmatrix}1 & 2 \\ 0 & 2 \end{bmatrix}$. Then the eigenvectors $v_1,v_2$ of $T$ are not orthogonal. If you use Gram Schmidt to orthogonalize $v_1,v_2$ one of them will no longer be an eigenvector.
If the resulting vectors were eigenvectors, then the matrix would have been orthogonally (or unitarily depending on the field) diagonalizable, which would imply that the matrix was normal. However, a quick check shows that $T T^* \neq T^* T$.
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
$v_3=u_3 - [(u_3|v_1)/(||v_1||^2)] \times v_1 - [(u_3|v_2)/(||v_2||^2)] \times v_2$
This is the Gram-Schmidt Orthogonalisation equation. If you apply this you would get answer straight away.
Since $u_3 \in \langle u_1,u_2\rangle$, Therefore, $u_3 \in \langle v_1,v_2\rangle$, because both $v_1=u_1 \; and \; v_2$ is a linear combination of $u_1,u_2$. Therefore $u_3 = a.v_1 + b.v_2$
Now $v_3 = a.v_1 + b.v_2 - (((a.v_1 + b.v_2)| v_1/||v_1||^2)\times v_1) + (((a.v_1 + b.v_2)| v_2/||v_2||^2)\times v_2) \\ = (a.v_1 + b.v_2) - (a.v_1 + b.v_2) = 0$
Remember that $(v_1|v_2) = 0$