[proofplan]
We prove measurability by testing all subsets of $G$ whose preimage under $f$ is Borel in $E$. This class is a $\sigma$-algebra because preimages commute with complements and countable unions. Continuity puts every open subset of $G$ in this class, so the class contains the $\sigma$-algebra generated by the open subsets of $G$, namely $\mathcal{B}(G)$. This is exactly the definition of $\mathcal{B}(E)/\mathcal{B}(G)$-measurability.
[/proofplan]
[step:Collect the subsets of $G$ whose preimages are Borel in $E$]
Let $\mathcal{P}(G)$ denote the power set of $G$. Define a collection $\mathcal{C}\subseteq \mathcal{P}(G)$ by
\begin{align*}
\mathcal{C}:=\{A\subseteq G:f^{-1}(A)\in \mathcal{B}(E)\}.
\end{align*}
We will prove that $\mathcal{C}$ is a $\sigma$-algebra on $G$ containing $\tau_G$.
[/step]
[step:Show the preimage class is a $\sigma$-algebra]
First, $G\in\mathcal{C}$ because
\begin{align*}
f^{-1}(G)=E\in\mathcal{B}(E).
\end{align*}
Let $A\in\mathcal{C}$. Then $f^{-1}(A)\in\mathcal{B}(E)$, and since $\mathcal{B}(E)$ is closed under complements in $E$,
\begin{align*}
f^{-1}(G\setminus A)=E\setminus f^{-1}(A)\in\mathcal{B}(E).
\end{align*}
Thus $G\setminus A\in\mathcal{C}$.
Now let $\mathbb{N}=\{1,2,3,\dots\}$ denote the set of positive integers, and let $(A_n)_{n\in\mathbb{N}}$ be a sequence in $\mathcal{C}$. For every $n\in\mathbb{N}$, $f^{-1}(A_n)\in\mathcal{B}(E)$, and since $\mathcal{B}(E)$ is closed under countable unions,
\begin{align*}
f^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)=\bigcup_{n=1}^{\infty}f^{-1}(A_n)\in\mathcal{B}(E).
\end{align*}
Therefore $\bigcup_{n=1}^{\infty}A_n\in\mathcal{C}$. Hence $\mathcal{C}$ is a $\sigma$-algebra on $G$.
[guided]
The purpose of introducing $\mathcal{C}$ is to isolate exactly the subsets of the codomain whose preimages already satisfy the desired measurability condition. Let $\mathcal{P}(G)$ denote the power set of $G$, and define
\begin{align*}
\mathcal{C}:=\{A\subseteq G:f^{-1}(A)\in \mathcal{B}(E)\}.
\end{align*}
To prove that $\mathcal{C}$ is a $\sigma$-algebra on $G$, we verify the three defining closure properties.
First, $\mathcal{C}$ contains the ambient set $G$. Indeed, the preimage of the whole codomain is the whole domain:
\begin{align*}
f^{-1}(G)=E.
\end{align*}
Since $\mathcal{B}(E)$ is a $\sigma$-algebra on $E$, it contains $E$. Hence $f^{-1}(G)\in\mathcal{B}(E)$, so $G\in\mathcal{C}$.
Second, $\mathcal{C}$ is closed under complements relative to $G$. Let $A\in\mathcal{C}$. By the definition of $\mathcal{C}$, this means $f^{-1}(A)\in\mathcal{B}(E)$. The preimage operation commutes with complements:
\begin{align*}
f^{-1}(G\setminus A)=E\setminus f^{-1}(A).
\end{align*}
Because $\mathcal{B}(E)$ is closed under complements relative to $E$, the set $E\setminus f^{-1}(A)$ belongs to $\mathcal{B}(E)$. Therefore $f^{-1}(G\setminus A)\in\mathcal{B}(E)$, which is exactly the statement that $G\setminus A\in\mathcal{C}$.
Third, $\mathcal{C}$ is closed under countable unions. Let $\mathbb{N}=\{1,2,3,\dots\}$ denote the set of positive integers, and let $(A_n)_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{C}$. For each $n\in\mathbb{N}$, the definition of $\mathcal{C}$ gives $f^{-1}(A_n)\in\mathcal{B}(E)$. Preimages commute with arbitrary unions, so in particular with countable unions:
\begin{align*}
f^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)=\bigcup_{n=1}^{\infty}f^{-1}(A_n).
\end{align*}
The right-hand side belongs to $\mathcal{B}(E)$ because $\mathcal{B}(E)$ is closed under countable unions. Hence
\begin{align*}
f^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)\in\mathcal{B}(E).
\end{align*}
By the definition of $\mathcal{C}$, this proves $\bigcup_{n=1}^{\infty}A_n\in\mathcal{C}$.
These three properties show that $\mathcal{C}$ is a $\sigma$-algebra on $G$.
[/guided]
[/step]
[step:Use continuity to include every open subset of $G$]
Let $U\in\tau_G$. Since $f:E\to G$ is continuous, the preimage $f^{-1}(U)$ is open in $E$, so $f^{-1}(U)\in\tau_E$. Because $\mathcal{B}(E)=\sigma(\tau_E)$ contains every member of $\tau_E$, we have $f^{-1}(U)\in\mathcal{B}(E)$. Therefore $U\in\mathcal{C}$, and hence
\begin{align*}
\tau_G\subseteq\mathcal{C}.
\end{align*}
[/step]
[step:Pass from open sets to all Borel subsets of $G$]
By the previous step, $\mathcal{C}$ is a $\sigma$-algebra on $G$ containing $\tau_G$. Since $\mathcal{B}(G)=\sigma(\tau_G)$ is the smallest $\sigma$-algebra on $G$ containing $\tau_G$, we obtain
\begin{align*}
\mathcal{B}(G)\subseteq\mathcal{C}.
\end{align*}
Thus, for every $B\in\mathcal{B}(G)$, the definition of $\mathcal{C}$ gives
\begin{align*}
f^{-1}(B)\in\mathcal{B}(E).
\end{align*}
This is precisely the definition of measurability of the map $f:(E,\mathcal{B}(E))\to(G,\mathcal{B}(G))$. Therefore $f$ is $\mathcal{B}(E)/\mathcal{B}(G)$-measurable.
[/step]
