In Andrew Baker's Matrix Groups, (in the proof of Theorem 20.11), there is an unproven statement that
if $G$ is a compact Lie group and $T$ is a maximal torus, then $\chi (G/T)\ne 0$.
I have an idea that $T \rightarrow G \rightarrow G/T$ is a fibration. How then can I use the maximality of $T$ to prove that $\chi(G/T)\ne 0$? Any hint please?
In fact a substantially stronger statement is true. $G/T$ is a smooth projective complex variety (called a complete flag variety). It has cohomology concentrated in even degrees because it admits an "affine cell decomposition," the Bruhat decomposition, and it's known that the cohomology of such a variety is free abelian on cells, where a cell of complex dimension $k$ corresponds to a cohomology class of degree $2k$.
In fact this decomposition has exactly $|W|$ cells, where $W$ is the Weyl group of the pair $(G, T)$, and hence we in fact have
$$\chi(G/T) = |W|.$$
For example, if $G = SU(2), T = U(1)$, then $G/T \cong \mathbb{P}^1$ has a Bruhat decomposition with two cells, namely a point and $\mathbb{A}^1$, and consequently $\chi(G/T) = 2$.
I don't know how to prove your desired statement without proving at least that the cohomology is concentrated in even degrees.