Product of a Continuous and Discrete Distribution 
Let $X \sim \mathcal{N}(0, 1)$ and $Y$ be a random variable independent of $X$ such that
  \begin{align*}
P(Y=y) = \begin{cases}
\frac{1}{2} & y = -1\\
\frac{1}{2} & y = 1\\
0 & otherwise
\end{cases}
\end{align*}
  If $ Z = XY$, show that $Z \sim \mathcal{N}(0,1)$.

Here is what I came up with:
\begin{align*}
P(Z<z) &= P(XY < z)\\
&=P(X < \frac{z}{Y})\\
&= P(X < \frac{z}{Y} \mid Y = -1)P(Y=-1) + P(X < \frac{z}{Y} \mid Y=1) P(Y=1)\\
&= \ldots\,?\\
&= P(X < z)
\end{align*}
Unfortunately, I'm not sure what to do, or whether I am even on the right track. I'd greatly appreciate any pointers!

Your mistake: you can't divide by $Y$ without knowing its sign (if it's negative, the inequality would reverse).
\begin{align*}
P(XY<z)
&= P(XY<z \mid Y=1) P(Y=1) + P(XY<z \mid Y=-1) P(Y=-1)\\
&=\frac{1}{2} P(X<z) + \frac{1}{2} P(-X<z)\\
&= \frac{1}{2} P(X<z) + \frac{1}{2} P(X>-z)\\
&= P(X<z).
\end{align*}
The last step is due to symmetry: $P(X<z)=P(X>-z)$
