$\mathbb{E}[f(X,Y)|Y]=\mathbb{E}[f(X,Y)]$ for all bounded measurable $f$ using the monotone class theorem

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Let $X,Y$ be independent rv's.
For $f:\mathbb{R}^2\to\mathbb{R}$ measurable we have $f_X:\mathbb{R}\to\mathbb{R}$ where $$f_X(y)= \left\{ \begin{array}{lr} \mathbb{E}[f(X,y)] & : \mathbb{E}|f(X,y)|<\infty\\ 0 & : \text{else} \end{array} \right..$$ Let $M_b$ be the set of all bounded measurable functions $\mathbb{R}^2\to\mathbb{R}$ and $$C=\{f\in M_b: \mathbb{E}[f(X,Y)|Y]=f_X(Y)\}.$$

How do we prove that $C=M_b$?

If we use the monotone class theorem, then for a $\pi-$system $\Pi$ that generates $\mathcal{B}^2$ we can show that $M_b\subset C$.
Therefore we want
1) $1_{\mathbb{R}^2}\in C$:
This is clear since for $A\in\sigma(Y)$ we have $$\int_A\mathbb{E}[1_{\mathbb{R}^2}(X,Y)|Y]d\mathbb{P}=\int_A1_{\mathbb{R}^2}(X,Y)d\mathbb{P}=\int_A1d\mathbb{P}=\int_A\mathbb{E}[1_{\mathbb{R}^2}(X,Y)]d\mathbb{P}.$$

2) for an increasing, non-negative sequence $(f_n)_n$ in $C$, where $f_n\to f$ with $f$ bounded, then $f\in C$:
So we want to show $$\mathbb{E}[f(X,Y)|Y]=\mathbb{E}[f(X,Y)]$$ Can we do this by using conditional monotone convergence and say $$\mathbb{E}[f_n(X,Y)|Y]\to\mathbb{E}[f(X,Y)|Y]$$ and since $$\mathbb{E}[f_n(X,Y)]\to\mathbb{E}[f(X,Y)]$$ by regular MCT, it follows?

3) for all $B\in \Pi$, $1_B\in C$:
Can we take $\Pi=\{(a,b)\times (c,d):a<b,c<d\in\mathbb{R}\}$? Then for $A\in\sigma(Y)$ $\begin{align*} \int_A\mathbb{E}[1_{(a,b)\times (c,d)}(X,Y)|Y]d\mathbb{P}&=\int_A1_{(a,b)\times (c,d)}(X,Y)d\mathbb{P}\\ &=\mathbb{P}[A\cap(X,Y)\in(a,b)\times (c,d)]\\ &=\mathbb{P}[A\cap\{X\in(a,b)\}]\mathbb{P}[A\cap\{Y\in (c,d)\}]\\ &=\int_A1_{X\in(a,b)}d\mathbb{P}\cdot\int_A1_{Y\in(c,d)}d\mathbb{P}\\ &=?\\ &=\int_A\mathbb{E}[1_{(a,b)\times (c,d)}(X,Y)]d\mathbb{P}. \end{align*}$
Is this correct? Which steps am I missing?