I am trying to attain the solution of Mathematical Statistics: Exercises and Solutions by Jun Shao Chapter $1$ Exercise $6$:
Let $f$ be a Borel function on $\mathcal{R}^2$. Define a function $g$ from $\mathcal{R}$ to $\mathcal{R}$ as $g(x)=f(x,y_0)$, where $y_0$ is a fixed point in $\mathcal{R}$. Show that $g$ is Borel. Is it true that $f$ is Borel from $\mathcal{R}^2$ to $\mathcal{R}$ if $f(x,y)$ with any fixed $y$ or fixed $x$ is Borel from $\mathcal{R}$ to $\mathcal{R}$?
Here is my attempt at solving this exercise:$$\text{For any fixed }y_0\text{, }\mathcal{G}\text{ is defined as}$$$$\mathcal{G}=\{C\subset\mathcal{R}^2:\{x:(x,y_0)\in C\}\in\mathcal{B}\}$$$$\emptyset\in\mathcal{G}\implies C\in\mathcal{G}\iff\{x:(x,y_0)\in C\}^c\in\mathcal{B}$$$$\implies C^c\in\mathcal{G}\iff C_i\in\mathcal{G}\text{, where }i=1\text{, }2\text{,}\dots$$$$\implies\{x:(x,y_0)\in\cup C_i\}=\cup\{x:(x,y_0)\in C_i\}\in\mathcal{B}\text{, }\cup C_i\in\mathcal{G}$$$$\therefore\mathcal{G}\text{ is a }\sigma\text{ field }\iff\text{ any open rectangle }(a,b)\times(c,d)\in\mathcal{G}$$$$\text{Which we know is true.}$$$$\text{And so, }\mathcal{G}\text{ contains }\mathcal{B}^2$$$$\text{Letting }B\in\mathcal{B}\text{, since }f\text{ is Borel, }A=f^{-1}(B)\in\mathcal{B}^2$$$$\therefore A\in\mathcal{G}\text{, and as such:}$$$$g^{-1}(B)=\{x:(x,y_0)\in A\}\in\mathcal{B}$$$$\text{This proves that }g\text{ is Borel.}$$$$\text{Now, if a function }f(x,y)\text{ with any fixed variable is Borel from }\mathcal{R}\text{ to }\mathcal{R}$$$$\text{the defined function will not necessarily be Borel from }\mathcal{R}^2\text{ to }\mathcal{R}$$$$\text{Let }A\text{ be a non-Borel subset of }\mathcal{R}\text{ where}$$ $$ f(x,y)= \begin{cases} 1 & x=y\in A \\ 0 & \text{otherwise} \\ \end{cases} $$ $$\text{For any function with a fixed point }f(x,y_0)=0\iff y_0\not\in A\text{ and }f(x,y_0)=I_{\{y_0\}}(x)\iff y_0\in A$$$$\therefore\text{ for any function with a fixed point, that function is Borel}$$$$\text{However, suppose the function is truly Borel. Then, }\mathcal{G}\text{ will be defined as}$$$$\mathcal{G}=\{C\subset\mathcal{R}^2:\{x:(x,x)\in C\}\in\mathcal{B}\}$$$$\text{If and when we use the same argument as in the first part of the proof,}$$$$\text{we should be able to show that }\mathcal{G}\text{ is a }\sigma\text{ field that contains }\mathcal{B}^2$$$$\therefore\{x:(x,x)\in B\}\in\mathcal{B}$$$$\text{However, by definition}$$$$\{x:(x,x)\in B\}=A\not\in\mathcal{B}$$$$\therefore\text{ }f(x,y)\text{ is not Borel}$$
$$\mathbf{\text{My question}}$$
Is the proof that I have formulated correct, or what could I do to prove it correctly/prove it more quickly?