Let us a consider a Lie group $G$, $S$ be a Lie subgroup and $EG$ the total space of the universal bundle $EG\twoheadrightarrow BG$. Let us consider the diagonal action $d$ of $G$ on the product $EG\times G/S$ and let us denote $EG\times_{G}G/S$ the quotient of $EG\times G/S$ through this action. I would like to show that $$ EG\times_{G}G/S=EG/S $$ where on $EG$ we consider the action of $S$ obtained restricting the action of $G$.
Intuitively I understand it but I'm having troubles in formalizing it. Indeed because $EG\twoheadrightarrow BG$ is a $G-$bunlde the action of $G$ preserves the fibers and acts on them freely and transitively. Considering now the product $EG\times G/S$, if for example we consider $g\in S$, then the action $d$ fixes $G/S$ and moves the elements of the fibers of $EG$. This in particular means that the cosets $gS$ move the elements of the product $EG\times G/S$ only long the fibers of $EG$. Hence this should imply that the orbit of the action of $G$ on a fiber $EG_{p}\simeq G$, where $p\in BG$, is split in a "number" of $G/S$ orbits. Thus, heuristically we get that $EG\times_{G}G/S=EG/S$. Do you have any hints to formalize this in a rigorous proof?