Your basic idea is right. However, you can easily verify that the vectors $u_1$ and $u_2$ you found are not orthogonal by calculating
$$<u_1,u_2> = (0,0,2,2)\cdot \left( \begin{matrix} 2 \\ 0 \\ -6 \\ -8 \end{matrix} \right) = -12-16 = -28 \neq 0$$
So something is going wrong in your process.
I suppose you want to use the Gram-Schmidt Algorithm to find the orthogonal basis. I think you skipped the normalization part of the algorithm because you only want an orthogonal basis, and not an orthonormal basis. However even if you don't want to have an orthonormal basis you have to take care about the normalization of your projections. If you only do $u_i<u_i,u_j>$ it will go wrong. Instead you need to normalize and take $u_i\frac{<u_i,u_j>}{<u_i,u_i>}$. If you do the normalization step of the Gram-Schmidt Algorithm, of course $<u_i,u_i>=1$ so it's usually left out. The Wikipedia article should clear it up quite well.
Update
Ok, you say that $v_1 = \left( \begin{matrix} 0 \\ 0 \\ 2 \\ 2 \end{matrix} \right), v_2 = \left( \begin{matrix} 2 \\ 0 \\ 2 \\ 0 \end{matrix} \right), v_3 = \left( \begin{matrix} 3 \\ 2 \\ -5 \\ -6 \end{matrix} \right)$ is the basis you start from.
As you did you can take the first vector $v_1$ as it is. So you first basis vector is $u_1 = v_1$ Now you want to calculate a vector $u_2$ that is orthogonal to this $u_1$. Gram Schmidt tells you that you receive such a vector by
$$u_2 = v_2 - \text{proj}_{u_1}(v_2)$$
And then a third vector $u_3$ orthogonal to both of them by
$$u_3 = v_3 - \text{proj}_{u_1}(v_3) - \text{proj}_{u_2}(v_3)$$
You did do this approach. What went wrong is your projection. You calculated it as
$$ \text{proj}_{u_1}(v_2) = v_2<u_1,v_2>$$
but this is incorrect. The true projection is
$$ \text{proj}_{u_1}(v_2) = v_2\frac{<u_1,v_2>}{<u_1,u_1>}$$
As I tried to point out, some textbooks will skip the division by $<u_1,u_1>$ in the explanation of Gram-Schmidt, but this is because in most cases you want to construct an orthonormal basis. In that case you normalize every $u_i$ before proceeding to the next step. Therefore $<u_i,u_i> = 1$ can be skipped.
So what you need to change is to divide by $<u_2,u_2> = 8$ in your projection.
Your answer is correct. Just note that in order for a set to form a basis, it must be linearly independent, and span the given space. It is easy to see in this example, but in further problems, you may need to verify manually.
Best Answer
To be orthogonal means $\langle u,v\rangle = 0$. In this case $\langle u,v\rangle=2(2 \cdot 1)-2\cdot 3-1 \cdot 1 + 1 \cdot 3=0$ so the are orthogonal with respect to this inner product.
A unit vector is a vector $x$ such that the norm (ie. length) $||x|| = \sqrt{\langle x,x\rangle} = 1$. For a set ot vectors to be orthonormal means they are all orthogonal to each other and all unit vectors. Again using the inner product given we can calculate that $\langle u,u \rangle = 2(1 \cdot 1)-3 \cdot 1 - 1 \cdot 3 + 3 \cdot 3 = 5 \neq 1$ so it's not a unit vector.
We can however find a unit vector in the same direction as $u$ if we scale $u$ by the reciprocal of it's length, so $(1/||u||)u= u /||u|| = u / \sqrt{5} = (1/\sqrt{5},3/\sqrt{5})$ and we can see that this is a unit vector because inner products are bilinear, so in particular for some scalar $k$ then $\langle kx,y\rangle = \langle x,ky\rangle = k\langle x,y\rangle$. In this case $\langle u/\sqrt{5},u / \sqrt{5} \rangle = (1/\sqrt{5})^2 \langle u,u \rangle = (1/\sqrt{5})^2 5 = 5 / 5 = 1$ so $u / ||u||$ is a unit vector. You can calculate this directly from the inner product formula as well. Similiarly $v / ||v||$ will be a unit vector, making $u / ||u||$ and $v / ||v||$ orthonormal with respect to this inner product.