I understand the classic version $E_X[X] = E_{Z}[E_{X|Z}[X|Z]]$.
My question relates to if $E_{(X,Y)}[XY] = E_X[E_{Y|X}[XY|X]]$ holds.
Suppose we are working with discrete random variables. My thought process:
$$ \begin{align} \text{E}_{(X,Y)}[XY]&= \sum_{x,y} x\text{ }y\text{ } \text{Pr}(X = x,Y = y)\\ &= \sum_{x,y} x\text{ }y\text{ } \text{Pr}(Y = y | X = x) \text{ }\text{Pr}(X = x)\\ &= \sum_{x,y}x \left( y \text{ }\text{Pr}(Y = y | X = x) \right)\text{ }\text{Pr}(X = x)\\ &= \sum_{x}x\left( \sum_{y }y \text{ }\text{Pr}(Y = y | X = x) \right)\text{Pr}(X = x)\\ &= \sum_{x}x \text{ }\text{E}_{Y|X=x}[Y|X=x]\text{ }\text{Pr}(X = x)\\ &= \sum_{x}x \text{ }g(x)\text{ }\text{Pr}(X = x)\\ &= E_X[g(X)]\quad (\text{where }g(t) = \text{E}_{Y|X=x}[Y|X=t)\\ &= E_X[\text{E}_{Y|X}[Y|X]] \end{align} $$
Is above correct? One can also derive in the similar way for continuous random variables right?
Your proof is almost completely correct. You just have a small mistake at the end. Recall, for any function $h$, that $$ \sum_x h(x)Pr(X=x)=E[h(X)] $$ Applying this to the function $h(x)=x\;g(x)$, your last couple of lines should be $$ \begin{align} E_{(X,Y)}[XY] &=\cdots \\&= \sum_x x \,g(x)\,Pr(X=x) \\&= E[Xg(X)] \\&= E[X\; E_{Y\mid X}[Y\mid X]] \\&= E[E_{Y\mid X}[XY\mid X]] \end{align} $$ The last equation is a basic property of conditional expectation. When conditioning on $X$, any function of $X$ behaves like a constant, so you can pull the $X$ inside the conditional expectation.
As you mentioned, the same proof works when $X,Y$ are jointly continuous, by replacing summations with integrals.