if $G$ is any compact group and $H$ is closed subgroup of $G$, then $G/H\rightarrow X_{H}\rightarrow X_{G}$ is a fibre bundle? ($X_G=X\times _{G}E=\left( X\times E\right) /G $ is orbit space where $X,G$-space and $G$ acts freely on $E$ contractible space and $G$ acts on $X\times E$ with $g\left( x,e\right) =\left( gx,ge\right) $)
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2$\begingroup$ You need to assume that $EG\to EG/G$ is a principal $G$--bundle, which doesn't follow from your assumptions (take $EG$ to be $G$ with the indiscrete topology to get a counterexample). Then you can take $EH = EG$, and your sequence is just the pullback of $G/H \to EG/H\to EG/G$ along the map $X_G \to EG/G$. $\endgroup$– Dan RamrasApr 8, 2016 at 19:44
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