Does $E(X|X\in A)=\frac{E(X\mathbf{1}(X\in A))}{Pr(X\in A)}$ hold? Does $E(X|X\in A)=\frac{E(X\mathbf{1}(X\in A))}{Pr(X\in A)}$ hold? (Here $\mathbf{1}(\cdot)$ is the indicator function). To me it seems that it holds. Here is the proof:
$E(X|X\in A)=\int_{-\infty}^{\infty}xf(x|x\in A)dx=\int_{-\infty}^{\infty}x\frac{f(x,x\in A)}{Pr(X\in A)}dx=\int_{-\infty}^{\infty}x\frac{f(x)\mathbf{1}(x\in A)}{Pr(X\in A)}dx=\frac{E(X\mathbf{1}(X\in A))}{Pr(X\in A)}$ .
Does this proof look correct?

Yes it does hold and your proof looks correct. You are simply using the definition of conditional probability, $f(x|x\in A)=\frac{f(x,x\in A)}{Pr(X\in A)}$.
