Suppose $Y \sim Ber(X)$ where $X = F(Z)$ and $Z \sim N(\mu, \sigma^2)$. What is the expected value $E[Y-X]$? Suppose $Y \sim Ber(X)$ where $X = F(Z)$ and $Z \sim N(\mu, \sigma^2)$. $F$ denotes the cdf of some continuous random variable.  What is the expected value $E[Y-X]$?
\begin{align*}
E[Y-X] &= E[Y]-E[X]\\
&= E[E[Y|X]] - E[X]\\
&= E[F(Z)] - E[F(Z)]\\
&= 0
\end{align*}
Is the above correct?

What you have done is correct (assuming that $F$ is fixed), but it can be simplified, generalised, and made more explicit.  Let's start by generalising your model and writing it in a more explicit form:
$$\begin{align}
Y|Z &\sim \text{Bern}(F(Z)), \\[6pt]
Z &\sim \text{Dist}, \\[6pt]
\end{align}$$
where $\text{Dist}$ is some arbitrary distribution.  Since $Y|Z \sim \text{Bern}(F(Z))$ we have the conditional expectation:
$$\mathbb{E}(Y|Z) = 0 \cdot (1-F(Z)) + 1 \cdot F(Z) = F(Z).$$
Using the law of total expectation we then have:
$$\begin{align}
\mathbb{E}(Y-F(Z))
&= \mathbb{E}(\mathbb{E}(Y-F(Z)|Z)) \\[6pt]
&= \mathbb{E}( F(Z) - F(Z) ) \\[6pt]
&= \mathbb{E}(0) \\[6pt]
&= 0. \\[6pt]
\end{align}$$
Observe that this result does not require any assumption about the distribution of $Z$.
