Let $X$ be a reasonable topological space (let's say it has the homotopy type of a CW complex) and let $G$ be a topological group acting on that space. Let $E_G \rightarrow B_G$ be the universal bundle. Then the equivariant cohomology $H^*_G(X)$ is defined as $H^*(E_G \times_G X)$. We call the space $E_G \times_G X$ the homotopy quotient space. We always have a map $E_G \times_G X \rightarrow G \backslash X$, where $G \backslash X$ denotes the naive quotient space. This induces a map $H^* (G\backslash X) \rightarrow H^*_G(X)$.

I'm curious about the following proof. From now on, we will take cohomology with rational coefficients. Suppose the action of $G$ on $X$ has finite stabilizers. Then the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is in fact an isomorphism. This is proved in the following notes by Michel Brion (http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf, see p. 4). The proof is as follows: the fibers of the map $E_G \times_G X \rightarrow G \backslash X$ are of the form $E_G/G_x$, where $G_x$ is the stabilizer of a point $x \in X$. Since $G_x$ is a discrete group, we have $\pi_1 (E_G/G_x) = G_x $ and vanishing of all higher homotopy groups. Since $G_x$ is finite, all the homotopy groups vanish when we tensor with $\mathbb{Q}$. I think this is what Brion means by "$\mathbb{Q}$-acyclic".

Then it is claimed that this is sufficient to prove that the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is an isomorphism. Why is this true? Could someone please explain this argument to me?

Even if the fibers were contractible, it seems to me that you would need this map to be a Serre fibration to prove a homotopy equivalence. Is it true that any surjective map with contractible fibers is a homotopy equivalence?