CDF of Piecewise Folded Normal I came across a problem in a Carmona's Statistical Analysis of Financial Data in R (pg. 189, Problem 3.13).  The due date has passed, so now it is considered a self-study question.

I am seeking a simple-to-understand proof to the following:
$X_1 \sim \mathcal{N}(0,1)$, $X_2 \sim \mathcal{N}(0,1)$
$X_3 \sim \begin{cases} 
      |X_2| & X_1 \geq 0 \\
      -|X_2| & X_1 < 0 
\end{cases} $
(Q.) Compute the CDF of $X_3$.  State if $X_3$ is Gaussian.

Intuitively, I believe $X_3$ should be standard normal $\sim \mathcal{N}(0,1)$.  Half of the time, $X_3 = $ folded normal, while half of the time $X_3 = $ negative of a folded normal, by simple symmetry.  I run an R simulation to test my suspicions and find this to be correct (the R code and plot is listed at the end of the question).
We know $\mathbb{P}(X_1 \leq 0) = \mathbb{P}(X_1 > 0) = \frac{1}{2}$ by symmetry.  We also know:
$\mathbb{P}(|X| \leq x) = \\
\mathbb{P}(-x \leq X \leq x) = \\
\mathbb{P}(X \leq x) - \mathbb{P}(X \leq -x) = \\
\mathbb{P}(X \leq x) - [1 - \mathbb{P}(X \leq x)] = \\
2 \cdot \mathbb{P}(X \leq x) - 1$
So, we try to say $X_3 \sim \frac{1}{2} \cdot [2 \cdot \mathbb{P}(X \leq x) - 1] + \frac{1}{2} \cdot [-2 \cdot \mathbb{P}(X \leq x) + 1]$, but the math here does not work out.  Any help from this point would be appreciated.

R Code and Result
# Variables
N <- 1000000
X1 <- rnorm(n=N, mean=0, sd=1); X2 <- rnorm(n=N, mean=0, sd=1)
X3 <- rep(0, N)

# Compute X3
for (i in 1:N) {
  if (X1[i] >= 0) {
    X3[i] <- abs(X2[i])
  } else {
    X3[i] <- -1*abs(X2[i])
  }
}

# Histogram (density)
hist(X3, freq=FALSE, breaks=50, 
     main="Histogram of X3 from Simulation",
     xlab="Value of X3",
     ylab="Density")
# Overlay standard normal for comparison
lines(density(X1), col="blue", lwd=2)


 A: I assume $X_1$ and $X_2$ are independent of each other in what follows (thanks for pointing out the importance of this go to @stubbornatom.)
I would work directly with the cumulative density function of $X_3$:
$$P(X_3 \leq x) = 0.5[P(|X_2|\leq x) + P(-|X_2| \leq x)]$$
Consider $x \leq 0$.  In this case, $P(|X_2| \leq x) = 0$, so, dropping it from the r.h.s. of the above equation, we have:
$$P(X_3 \leq x) = 0.5P(-|X_2| \leq x) = 0.5[P(X_2 \leq x) + P(-X_2 \leq x)]$$
where the second term on the r.h.s. covers the two cases $X_2 \leq 0$ (in which case $-|X_2| = X_2$) and $X_2 > 0$ (in which case $-|X_2|=-X_2$) respectively.   Multiplying the terms of the last probability by $-1$ and switching the direction of the inequality results in:
$$P(X_3 \leq x) = 0.5[P(X_2 \leq x) + P(X_2 \geq -x)] = P(X_2 \leq x)$$
where the last equality is due to the symmetry of the standard Normal distribution around $0$.
A similar logic can be used to show that $P(X_3 \leq x) = P(X_2 \leq x)$ for the case $x > 0$.  (Actually I think saying "symmetry" is probably sufficient.) 
 Since the CDF of $X_3$ is everywhere equal to the CDF of $X_2$, and $X_2$ is standard Normal, it follows that $X_3$ is standard Normal too.
