Here's the argument I know that avoids spectral sequences, based on the
little-known space $G/N(T)$.

In between $T$ and $G$ is $N(T)$. Note that $EG$ "is an" $ET$ and $EN(T)$,
since it's contractible and $T,N(T)$ act freely on it, so we can
identify $BT, BN(T)$ with $EG/T, EG/N(T)$.

Now consider the two maps $EG/T \to EG/N(T) \to EG/G$, with fibers
$W = N(T)/T$ and $G/N(T)$ respectively. The first case divides by a
free action of $W$, so we can identify
$H^\ast(BN(T);{\mathbb Q}) = H^\ast(BT; {\mathbb Q})^W$ by pushing and pulling.
(Actually we only need to invert $|W|$, and generally less; for $G=U(n)$
it's true over $\mathbb Z$.) In particular, there is only
even cohomology.

So let's look at the space $G/N(T) = (G/T)/W$. The space $G/T$ has a Bruhat
decomposition, hence only even-degree cohomology (even over $\mathbb Z$),
which you can prove via Morse theory on a generic adjoint orbit if you don't
want to bring in algebraic geometry, and its Euler characteristic is $|W|$.

Hence the space $(G/T)/W$ has (rationally) only even-degree cohomology,
and Euler characteristic $1$. So it has the rational cohomology of a point!
For $G=SU(2)$ this space is ${\mathbb RP}^2$.

By a particularly trivial application of Leray-Hirsch (which I think
is the only remainder of the spectral sequence argument Mark Grant gave),
$H^\ast(EG/G; {\mathbb Q}) \cong H^\ast(EG/N(T); {\mathbb Q})$.

(Oops: I guess this answer isn't so different from Ralph's.)

viz, e.g.Figueroa-O'Farrill's answer. – Dr Shello Apr 17 '11 at 7:48