Is this proof of the Law of Total Probability correct? I start with the Law:
$$\operatorname{P}(X_1 \le x_1) = \operatorname{E} \bigl( \operatorname{P}(X_1 \le x_1 \vert X_2) \bigr)$$
And I express the expectation as:
$$\begin{align}
\operatorname{E} \bigl( \operatorname{P}(X_1 \le x_1 \vert X_2) \bigr) = & \int_{-\infty}^{\infty} \operatorname{P}(X_1 \le x_1 \vert X_2 = x_2)\;f_2(x_2)\;dx_2 \\
& = \int_{-\infty}^{\infty} \Biggl(\int_{t \le x_1} \frac{f_{X_1,X_2}(t,x_2)}{f_{X_2}(x_2)} \; dt \Biggr)\;f_{X_2}(x_2)\;dx_2 \\
& = \int_{-\infty}^{\infty} \Biggl(\int_{-\infty}^{x_1} \frac{f_{X_1,X_2}(t,x_2)}{f_{X_2}(x_2)} \; dt\Biggr)\;f_{X_2}(x_2)\;dx_2 \\
& = \int_{-\infty}^{\infty} \int_{-\infty}^{x_1} \frac{f_{X_1,X_2}(t,x_2)}{f_{X_2}(x_2)}\;f_{X_2}(x_2) \; dt \; dx_2 \\
& = \int_{-\infty}^{\infty} \int_{-\infty}^{x_1} f_{X_1,X_2}(t,x_2) \; dt \; dx_2 \\
& = \int_{-\infty}^{x_1} \int_{-\infty}^{\infty} f_{X_1,X_2}(t,x_2) \; dx_2 \; dt \\
& = \int_{-\infty}^{x_1} f_{1}(t) \; dt \\
& = \operatorname{F}_{X_1}(x_1) = \text{P}(X_1 \le x_1)
\end{align}$$
I am especially not sure about changing the integrals around, as in a previous course I was told that the integral with the variable in the bounds had to be integrated first.
 A: Well, firstly $\textsf{P}(X_1 \le x_1 \mid X_2 = x_2)$ is a constant, not a random variable, but that is an easy fix.   $\textsf{P}(X_1 \le x_1 \mid X_2)$ is a random variable with the requisite properties.
Secondly, $x_1$ is not a variable of integration, so presents no problem for the change of order; and the integration order can be changed via Fubini/Tonelli's Theroem.   I suggest avoid invoking $x_2$ and rather using both $s,t$ as the variables of integration to be clear about the distinction.
Finally, watch the capitalisation for the subscripts of the density functions; it is case sensitive.
These issues aside, your work was otherwise okay.
$$\begin{align}
\mathsf{E} \bigl( \mathsf{P}(X_1 \le x_1 \mid X_2) \bigr) = & \int_{-\infty}^{\infty} \mathsf{P}(X_1 \le x_1 \mid X_2 = s)\;f_{\small X_2}(s)\;ds \\
& = \int_{-\infty}^{\infty} \Biggl(\int_{t \le x_1} \frac{f_{\small X_1,X_2}(t,s)}{f_{\small X_2}(s)} \; dt \Biggr)\;f_{\small X_2}(s)\;ds \\
& = \int_{-\infty}^{\infty} \Biggl(\int_{-\infty}^{x_1} \frac{f_{\small X_1,X_2}(t,s)}{f_{\small X_2}(s)} \; dt\Biggr)\;f_{\small X_2}(s)\;ds \\
& = \int_{-\infty}^{\infty} \int_{-\infty}^{x_1} \frac{f_{\small X_1,X_2}(t,s)}{f_{\small X_2}(s)}\;f_{\small X_2}(s) \; dt \; ds \\
& = \int_{-\infty}^{\infty} \int_{-\infty}^{x_1} f_{\small X_1,X_2}(t,s) \; dt \; ds \\
& = \int_{-\infty}^{x_1} \int_{-\infty}^{\infty} f_{\small X_1,X_2}(t,s) \; ds \; dt \\
& = \int_{-\infty}^{x_1} f_{\small X_1}(t) \; dt \\
& = F_{\small X_1}(x_1) \\ &= \mathsf{P}(X_1 \le x_1)
\end{align}$$
