Let $G$ be an affine, complex, reductive algebraic group and $B$ a Borel of $G$. I have seen and understood the proof that $G/B$ is projective. Now, on the other hand, I have made the following reasoning which must be false: Let $U$ be the unipotent radical of $B$ and $T$ a maximal torus of $G$ such that $B=UT$. Then, $\newcommand{\C}{\mathbb{C}}\C[G]^U$ is a finitely generated $\C$-algebra. Hence, $G/U$ is an affine variety. On this affine variety, we have an action of the torus $T$, which is affine and reductive. The (categorical) quotient of an affine variety by an affine reductive group is affine, so $$\C[(G/U)/\!/T] =\C[G/U]^T=\C[G]^{UT}=\C[G]^B$$ is finitely generated. This seems to imply that $G$ has an affine, categorical quotient by $B$. But since geometric quotients are also categorical quotients and those are unique, $G/B$ would be affine and that's not true.
Where exactly is the mistake?
$G/U$ is not necessarily an affine variety. Take for example $G = SL_2(\mathbb{C})$ and $U = \left\{ \begin{pmatrix} 1 & \ast \\ 0 & 1 \end{pmatrix} \right\}$. Then $G/ U \cong \mathbb C^{2} \setminus 0$ because $U$ is the stabiliser of $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ with respect to the transitive (linear) action of $G$ on $\mathbb{C}^{2} \setminus \{0\}$.
But yes, $\mathbb{C}[G/U] = \mathbb{C}[x,y]$ is finitely generated in this case, too.