We know that $\mathbb{E}(XY|X)$ is itself a random variable with respect to $x$. So, to prove the given statement in the title, it suffices to treat $\mathbb{E}(XY|X)$ as a function of $x$ and prove that it holds for any $x$. Hence
$$\mathbb{E}(XY|X)(x)=\mathbb{E}(XY|X=x)=\mathbb{E}(xY|X=x)=x\mathbb{E}(Y|X=x)=\left( X\mathbb{E}(Y|X) \right)(x)$$
Since that holds for any $x$, we conclude that $\mathbb{E}(XY|X)=X\mathbb{E}(Y|X)$. Q.E.D.
This fact, together with the law of total expectation, proves that we have
$$\mathbb{E}(XY)=\mathbb{E}(\mathbb{E}(XY|X))=\mathbb{E}(X\mathbb{E}(Y|X))$$
Am I correct?
It depends on your definition of conditional expectation. A standard definition of conditional expectation is given here
Now observe that this is equivalently stated as, for any $\Sigma$-measurable function $f$, $\mathbb{E}[X|Y]$ is the only function $g$ of $Y$ that satisfy $$\mathbb{E}[g(Y) f(Y)]=\mathbb{E}[Xf(Y)]$$
In your case, $\mathbb{E}[Y|X]$ is a function of $X$ and so $X\mathbb{E}[Y|X]$ too. For any $\Sigma$-measurable function $f$, let $f'(x)=xf(x)$, this is $\Sigma$-measurable, so we have \begin{align} \mathbb{E}[X\mathbb{E}[Y|X]f(X)]&=\mathbb{E}[\mathbb{E}[Y|X](X f(X))]\nonumber\\ &\stackrel{(\ast)}{=}\mathbb{E}[Y(X f(X))]\\ &=\mathbb{E}[XY f(X)]\nonumber\\ \end{align}
Where $(\ast)$ follows by definition of the conditional expectation $\mathbb{E}[Y|X]$. This proves that $\mathbb{E}[XY|X]=X\mathbb{E}[Y|X]$.