Homotopy fixed point on the fibre of a fibration over BG I am currently reading the paper Homotopy Fixed Point Methods for Lie Groups and Finite Loop Spaces by Dwyer and Wilkerson (can be found here http://www.math.purdue.edu/~wilker/papers/analysis.pdf for example), and I'm having trouble proving a basic result in the appendix on homotopy fixed points, and any help would be much appreciated. 
Assumption: All groups are discrete groups, and $EG$ and $BG$ are CW-complexes (for example coming from the usual nerve construction).

Lemma 10.4: Let $W \rightarrow BG$ be a (Serre) fibration with fibre $F$. Then there is
  a proxy action of $G$ on $F$ such that $F^{hG}$ is homotopy equivalent to the
  space of sections of $W \rightarrow BG$.

Here proxy action (as they defined it a couple of paragraphs before the statement of the lemma) just means that there is an ordinary homotopy equivalence $F \simeq X$ where $X$ is a $G$-space and $F^{hG} := X^{hG}$. 
Following their proof, we're supposed to consider the pullback $\tilde{W}$ for the proxy action space 
$\require{AMScd}$
\begin{CD}
\tilde{W} @>{h}>> W\\
@VVV @VV{p}V \\
EG @>{q}>> BG
\end{CD}
and since $EG$ is contractible, we obtain that $\tilde{W} \simeq F$. Now we need to show that $\tilde{W}$ has the correct homotopy fixed point as required. I've tried two approaches that I can't seem to make work.
Method 1: Recall now that $\tilde{W}^{hG} := Map^G(EG,\tilde{W})$ the space of $G$-equivariant maps, and write $\Gamma(W\xrightarrow{p} BG)$ for the space of sections. So a point in $\tilde{W}^{hG}$ is just a commuting square  $\require{AMScd}$
\begin{CD}
EG @>{f}>> W\\
@V{a}VV @VV{p}V \\
EG @>{q}>> BG
\end{CD}
where $a$ is a $G$-map and a point in $\Gamma(W\xrightarrow{p} BG)$ is just a commuting square
$\require{AMScd}$
\begin{CD}
EG @>{s}>> W\\
@V{id}VV @VV{p}V \\
EG @>{q}>> BG
\end{CD}
So we can clearly get a map $\Gamma(W\xrightarrow{p} BG) \rightarrow \tilde{W}^{hG} $. 
To get a map the other way, recall that the $G$-Whitehead theorem (for instance 6.4.2 of Benson's Representations and Cohomology V2) gives that any $G$-self-map between $EG$ is a $G$-homotopy equivalence. So we obtain a diagram $\require{AMScd}$
\begin{CD}
EG @>{\bar{a}}>> EG @>{f}>> W\\
@V{id}VV@V{a}VV @VV{p}V \\
EG @>{id}>> EG @>{q}>> BG
\end{CD} 
where the left square commutes up to a $G$-homotopy, call it $\gamma$ say. Using this homotopy, that $p$ was a fibration, and looking at the diagram $\require{AMScd}$
\begin{CD}
EG\times 0 @>{f}>> W\\
@VVV @VV{p}V \\
EG\times I @>>{\gamma}> BG
\end{CD}
we can get a point in $\Gamma(W\xrightarrow{p} BG)$. My problem with this is that I don't know how to show that the two maps so defined give the required homotopy equivalence. In fact I have some doubts that this will work since I've made so many choices in the construction of the second map.
Method 2: This uses the usual identification of $\tilde{W}^{hG}$ with the space of sections of the canonical fibration $\tilde{W}_{hG} \xrightarrow{\pi} BG$ where $\tilde{W}_{hG} := EG \times_G \tilde{W}$ the homotopy orbit space/Borel construction. One can show that the natural map $\tilde{W}_{hG} \xrightarrow{\bar{h}} W$ is a homotopy equivalence. Then I sort of want to say that because of this homotopy equivalence we obtain $\Gamma(\tilde{W}_{hG} \xrightarrow{\pi} BG) \simeq \Gamma(W \xrightarrow{p} BG)$, but the problem with this is that the two fibrations are using different copies of $EG$.
 A: In case anyone's interested, here's a solution due to a couple of friends.
Write $\tilde{W}^{hG} = Map^G(EG,\tilde{W}) = Map^G(EG,EG)\times_{Map^G(EG,BG)}Map^G(EG,W)$ and $\Gamma(W\rightarrow BG) = *\times_{Map^G(EG,BG)}Map^G(EG,W)$.

Claim: $EG \xrightarrow{\simeq} *$ induces $Map^G(EG,EG) = (EG)^{hG} \xrightarrow{\simeq} (*)^{hG} = *$, that is, that  $Map^G(EG,EG)$ is contractible.

Given the claim, we then look at the diagram
$\require{AMScd}$
\begin{CD}
* @>>> Map^G(EG,BG) @<<< Map^G(EG,W)\\
@V{\simeq}VV @V{id}VV @V{id}VV\\ 
Map^G(EG,EG) @>>> Map^G(EG,BG) @<<< Map^G(EG,W)\\
\end{CD}
and use for example (the dual of) Hovey's cube lemma (Lemma 5.2.6 of https://web.math.rochester.edu/people/faculty/doug/otherpapers/hovey-model-cats.pdf), together with the fact that any space is fibrant and that the right horizontal maps are fibrations to get the required homotopy equivalence of pullbacks.
Proof of claim: The unique $G$-map $EG \xrightarrow{\simeq} *$ induces a $G$-map $$EG\times EG \xrightarrow{pr_1} EG $$ between free $G$-spaces, so by $G$-Whitehead, is a $G$-homotopy equivalence. So quotienting out by $G$ we obtain $$(EG)_{hG} \xrightarrow{\simeq} (*)_{hG} = BG$$  We know that in fact this gives a fibre homotopy equivalence 
$\require{AMScd}$
\begin{CD}
(EG)_{hG} @>{\simeq}>> (*)_{hG} = BG\\
@VVV @VVV\\
BG @>{id}>> BG \\
\end{CD}
for example from page 52 of May's Concise Algebraic Topology https://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf. But then the space of sections of the two vertical fibrations are homotopy equivalent, and we know that homotopy fixed point spaces can be identified with such spaces of sections. $\square$
Comment: The result is quite interesting as it provides sort of a converse to the well-known fact that for any $G$-space $X$ the homotopy fixed points $X^{hG}$ can be identified with the space of sections to the fibration $X_{hG} \rightarrow BG$.
