I would like a proof (or a sketch of a proof or a reference) for the following fact:
If $\pi \colon P \to M$ is a fiber bundle of manifolds, and its fibers are contractible, then there exist a continuous sections.
If necessary you may suppose the fibers are discs or that the base space is compact.
The reason why I'm asking this question is that this fact is used in paper Representations of surface groups in complex hyperbolic space - D. Toledo to show the existence of equivariant maps.
Is there a geometrical reason for this fact to be true? I would like to have an intuition of this.
If the base space $M$ is a CW complex then the problem of Existence Of A Section is solved using Obstruction Theory. (Steenrod's book "The Topology of Fibre Bundles" gives a great exposition in the classical "cell-by-cell" case, and you can find a more modern approach using Postnikov towers in Hatcher.)
(Edit: See Tyrone's comment for an alternate approach which doesn't use CW structures.)
The very rough idea is that if $\sigma\colon M^{(n)} \to P$ is a section on the $n$-skeleton and $D$ is an $(n+1)$-cell then $\sigma$ can be extended over $D$ iff $\sigma|_{\partial D}$ is null-homotopic, and since $D$ is contractible we can push the problem into a fibre over a point $x\in D$ and get a homotopy class $[\sigma_D] \in \pi_n(F_x)$ such that $\sigma$ can be extended over $D$ iff $[\sigma_D] = 0$. Since $F$ is contractible this homotopy class vanishes, and we can always extend the section.
To finish the obstruction theory picture, as $D$ varies over the $(n+1)$-cells all of these homotopy classes incredibly form an "obstruction class"
$$\mathfrak{o}(\sigma)\in H^{n+1}(M;\{\pi_n(F_x)\})$$
where $\{\pi_n(F_x)\}$ is a "bundle of coefficients" (if $M$ is simply-connected then the coefficients are just $\pi_n(F)$). Then $\sigma$ can be extended to the $(n+1)$-skeleton if and only if $\mathfrak{o}(\sigma)$ vanishes. If $F$ is contractible then all of these groups vanish, so a global section can be inductively constructed on the skeleta of $M$.