Normal distribution, uncorrelated random variables $Y$ is a random variable such that $$Y = \cases{X,\hspace{2mm} \mathrm{for}\hspace{1mm} |X| \leq c \\ \\-X,\hspace{2mm}\mathrm{for}\hspace{1mm}|X|>c,} $$ where $c \geq 0,$ and $X$ has the standard normal distribution. I'd like to show that there exists a $c$ so that $\mathrm{Cov}(X,Y)=0$. 
What I know so far is that $Y$ also has the standard normal distribution, which means that $\mathrm{E}X \mathrm{E}Y = 0 \cdot 0.$ Hence, I suppose I need to find some $c$ that will make $\mathrm{E}XY$ equal to zero, in order to get $\mathrm{Cov}(X,Y) = \mathrm{E}XY - \mathrm{E}X\mathrm{E}Y = 0$. I haven't had any success, though.
Since if $c \rightarrow \infty$, we get $\mathrm{Cov}(X,Y) = 1$, I suppose that if $c \rightarrow 0$, we'll have $\mathrm{Cov}(X,Y) = -1$ (?), but I don't know if that is something I can use. I'm pretty lost on how to solve this. 
 A: $$XY = \begin{cases}X^2 & |X| \le c \\ -X^2 & |X|>c\end{cases}$$
$$E[XY] = E[X^2 \mathbf{1}_{|X| \le c}] - E[X^2 \mathbf{1}_{|X| > c}]$$
(I think you already did the above.)
So as you noted, when $c \to \infty$ this becomes approaches $1$, while when $c \to 0$ this approaches $-1$. If you accept that the above quantity varies continuously with $c$, then you are done (intermediate value theorem). I think proving continuity is pretty straightforward; you can write down the integral and even use integration by parts to give an explicit expression in terms of $c$ and the CDF of the standard normal.
A: You are correct that $Y \sim \text{N}(0,1)$ under your specification, so you have $\mathbb{E}(X) = \mathbb{E}(Y) = 0$.  Letting $y(x)$ be your specified function and letting $\phi$ be the standard normal density you then have:
$$\begin{equation} \begin{aligned}
\mathbb{Cov}(X,Y) = \mathbb{E}(XY) &= \int \limits_{-\infty}^\infty x y(x) \phi(x) dx \\[8pt]
&= - \int \limits_{-\infty}^c x^2 \phi(x) dx +
 \int \limits_{-c}^c x^2 \phi(x) dx -\int \limits_c^\infty x^2 \phi (x) dx \\[8pt]
&= 4 \int \limits_0^c x^2 \phi (x) dx  - \int \limits_{-\infty}^\infty x^2 \phi(x) dx \\[8pt]
&= 4 \int \limits_0^c x^2 \phi (x) dx - 1. \\[8pt]
\end{aligned} \end{equation}$$
Now, using integration by parts, it can be shown that:
$$I(c) \equiv \int \limits_0^c x^2 \phi(x) dx = \int \limits_0^c \Big( \phi(x) - \phi (c) \Big) dx.$$
Hence, we have $\mathbb{Cov}(X,Y) = 0$ at the point where:
$$\begin{equation} \begin{aligned}
\int \limits_0^c \Big( \phi(x) - \phi (c) \Big) dx = \frac{1}{4}.
\end{aligned} \end{equation}$$
This integral condition can be solved numerically to yield $c = 1.538172$ (rounded to six decimal places).  Note that with this value the random variables $X$ and $Y$ are uncorrelated, but of course, they are not independent.
