Why universal G-bundles are contractible?

Let $G$ be a nice topological group and $E\to B$ a universal $G$-bundle. I'm interested in a proof of contractibility of $E$ using only the universal property of it. I also know that if there is a contractible $G$-bundle, then all of others are also contractible, but does there exist a direct proof not using a special construction of contractible universal $G$-bundles?

• If $B$ represents the functor «$X$ → principal $G$-bundles on $X$», then $E$ represents the functor «$X$ → principal $G$-bundles on $X$ with a fixed section». This functor is trivial, so $E$ is contractible. – Grigory M Jun 4 '15 at 21:57
• @GrigoryM I know that a map to $E$ gives a section of a $G$-bundle but I'm having trouble to prove your correspondence for homotopy classes of maps. Could you please write a complete answer? thanks! – Mostafa Jun 5 '15 at 18:43
• Nobody has an answer for this important fact? – Eduardo Longa Jun 10 at 18:49

First note that the projection $$\pi: E \rightarrow B$$ is nullhomotopic by the universal property of $$B$$ since the pullback of a principal bundle by its projection is always trivial.
It follows from clutching and the loop-suspension adjunction that both $$\Omega B$$ and $$G$$ represent $$G$$-bundles over the suspension of your space.
This means that we can use the following rectangle which is commutative up to homotopy (formatting taken from https://mathoverflow.net/a/132340/134512) . $$\begin{matrix} G & \to & E & \to & B \\ \downarrow \simeq && \downarrow * && \downarrow = \\ \Omega B & \to & PB & \to & B \end{matrix}$$
This square induces a morphism of long exact sequences, which by the five lemma is an isomorphism from the homotopy groups of $$B$$ to the trivial group, as desired.