How do I show (using properties of conditional expectation) that the following expression is equal to 0?
$$\begin{align} \mathbb{E}[\mathbb{E}[g(X)] - g(X)|X] &= \mathbb{E}[\mathbb{E}[g(X)]|X] - \underbrace{\mathbb{E}[g(X)|X]}_{g(X)\mathbb{E}[1|X] = g(X)}\\ \end{align}$$
This is what I have so far. I can't figure out how to handle the first term.
$\mathsf E[g(X)]$ is a constant, so can be distributed out of the linear operation (expectation).
$$\mathsf E[\mathsf E[g(X)]\mid X]~{~=~\mathsf E[g(X)]\,\mathsf E[1\mid X]\\ ~=~ \mathsf E[g(X)]}$$
On the other hand, $g(X)$ is $X$ measurable, so the conditional expectation for it when given $X$ is itself. $$\mathsf E[g(X)\mid X]~{~=~g(X)\,\mathsf E[1\mid X]\\~=~g(X)}$$
(We normally skip passed those middle steps and go straight to goal, but since you put it in the underbrace, I'll just confirm that it is okay.)
Putting it together:
$$\begin{align} \mathsf{E}[\mathsf{E}[g(X)] - g(X)\mid X] ~&=~ \mathsf{E}[\mathsf{E}[g(X)]\mid X] - \mathsf{E}[g(X)\mid X]\\ &=~ \mathsf E[g(X)]-g(X) \end{align}$$
This is not guaranteed to equal zero; although its expectation obviously is.
$$\begin{align} \mathsf E[\mathsf{E}[\mathsf{E}[g(X)] - g(X)\mid X]] ~&=~ \mathsf{E}[\mathsf E[g(X)]-g(X)] \\ &=~ 0 \end{align}$$