Define a bivariate random variable $(X, Y)$ as follows. Let joint pdf be $f_{X,Y}(x, y) = (1/\pi)\exp\{(-1/2)(x^2+y^2)\}$ if ($x > 0$ and $y > 0$) or ($x < 0$ and $y < 0$) and $f_{X,Y}(x, y) = 0$ otherwise.
Need to show that $X + Y$ is not normal without finding the exact distribution of the sum.
I showed that both $X$ and $Y$ are normal distributed with parameters $(0, 1).$ It's clear that $X$ and $Y$ should be dependent random variables, then sum $X + Y$ is not normal distributed. I thought about considering a case when $Y = |X|$ but I'm not sure how to show that $X + Y$ is not normal in this case.
Clearly $X+Y$ has mean $0$. If $X+Y\sim N(0,\,\sigma^2)$ then $(X+Y)^2,\,(X+Y)^4$ have respective means $\sigma^2,\,3\sigma^4$, i.e.$$E[(X+Y)^4]=3(E[(X+Y)^2])^2.$$So, let's refute that. Write $\mu_n:=\int_0^\infty x^ne^{-x^2/2}dx$: we'll repeatedly use$$\mu_0=\sqrt{\frac{\pi}{2}},\,\mu_1=1,\,\mu_2=\sqrt{\frac{\pi}{2}},\,\mu_3=2,\,\mu_4=3\sqrt{\frac{\pi}{2}}.$$Letting $S:=[0,\,\infty)^2\cup(-\infty,\,0]^2$ denote the support of $(X,\,Y)$,$$\begin{align}E[(X+Y)^2]&=\int_S\frac{1}{\pi}(x^2+y^2+2xy)e^{-(x^2+y^2)/2}dxdy\\&=\int_{[0,\,\infty)^2}\frac{2}{\pi}(x^2+y^2+2xy)e^{-(x^2+y^2)/2}dxdy\\&=\frac{4}{\pi}(\mu_0\mu_2+\mu_1^2)\\&=\frac{4}{\pi}+2.\end{align}$$Similarly,$$E[(X+Y)^4]=\frac{4}{\pi}(\mu_0\mu_4+4\mu_1\mu_3+3\mu_2^2)=\frac{32}{\pi}+12\ne3\left(\frac{4}{\pi}+2\right)^2.$$In particular, $22.2\ne32.1$ (or you could use the fact that $\pi$ is transcendental).