Let $G$ be a compact connected Lie group, $T$ be some maximal torus in $G$ (that is, inclusionmaximal connected abelian subgroup). Then the union of tori $gTg^{1}$, $g\in G$, is the whole $G$. This is wellknown (4.21 in Adams book). My question is rather methodological: is there any proof without use of algebraic topology? Adams presents A. Weil's proof, which uses some kind of Lefschetz fixed points theorem. (Yes, I am sorry but my motivation is mostly that I teach second year students this stuff.)

There are proofs that avoid algebraic topology: see for example Chapter 16 in Bump's Lie Groups or IV.5 in Knapp's Lie Groups Beyond an Introduction. 


There are many ways of proving this, using all sorts of different methods. In the second edition my book "Lie groups, Lie algebras, and representations" (following Brocker and tom Dieck) I use the mapping degree theorem. If we fix a single maximal torus $T$ and consider the conjugation map $\Phi:T \times (K/T) \rightarrow K$ be given by $\Phi(t,[x])=xtx^{1}$. If we can show that this map has nonzero mapping degree, we can conclude it is surjective, which is just what we are trying to show. In Section 11.5, I show that $\Phi$ has mapping degree equal to the order of the Weyl group. This approach also gives a proof of the Weyl integral formula by the same computation. 


The claim in the question that maximal tori are the same as inclusionmaximal abelian subgroups is not correct. For example, the diagonal matrices with +1 or 1 on the diagonal form a maximal abelian subgroup of SO(n) that is not a torus. 

