Let $X,Y \in L^2$ be independent random variables and $G$ be a sub $\sigma$-algebra. Then $$E[XY|G]=E[X|G]\cdot E[Y|G].$$
Here is my alleged proof: Since $X,Y \in L^2$, all the conditional expectations are well-defined. Then $$ \begin{align}E[XY|G]&=E\bigg[E[XY|\sigma(G,X)]\big|G\bigg]\ \ \ \ (\because\text{tower property}) \\&=E\bigg[X\ E[Y|\sigma(G,X)]\big|G\bigg] \\&=E\bigg[X\ E[Y|G]\big|G\bigg] \ \ \ \ (\because \text{$X$ and $Y$ indep}) \\&=E[Y|G]\cdot E[X|G] \ \ \ (\because \text{$E[Y|G]$ is $G$ measurable}) \end{align}$$ yields the result.
Is my proof okay? Any help is appreciated.
As I commented, there is a gap in the second to the last step; we need that $X$ is independent with $\sigma(Y,G)$.
Let $\varepsilon_1$ and $\varepsilon_2$ be two independent identically distributed random variables taking the values $-1$ and $1$ with probability $1/2$. Let $X=\varepsilon_1\varepsilon_2$, $Y=\varepsilon_1$ and $\mathcal G=\sigma(\varepsilon_2)$. Then $X$ is independent of $Y$, $$ \mathbb E\left[XY\mid \mathcal G\right]=\mathbb E\left[ \varepsilon_2\mid \mathcal G\right]=\varepsilon_2 $$ and $$ \mathbb E\left[Y\mid \mathcal G\right]=0. $$ Therefore, it seems that we cannot avoid the independence of $X$ with $\sigma(\mathcal G,Y)$ or a similar assumption by interchanging $X$ and $Y$.