Let $(X_1, X_2)$ be a pair of uniform distributed random coordinates on $S:=([-1,0]\times[-1,0])\;\cup\; [0,1]\times[0,1])$. Are $X_1, X_2$ independent? Are $X_1, X_2$ uncorrelated?
If you draw the picture you can see they're obviously dependent are positively correlated. If I know $X_1$ is negative, then by all means $X_2$ is negative as well, and vice versa. Also, they occur in the same "direction". So positively correlated. The intuition is quite obvious. Here is my formal approach:
Show that $f_{X,Y}(x,y)\not= f_X(x)f_Y(y)$.
We know that $$f_{X,Y}(x,y)=\frac{1}{2}\left(\mathbf{1}_{[-1,0]^2}(x,y)+\mathbf{1}_{[0,1]^2}(x,y)\right)$$ and we use $1/2$ because when we integrate the above it needs to equal 1. To obtain $f_X(x)$ we integrate over $y$ and analogous for $f_Y(y)$. $$f_X(x)=\int_{\mathbb{R}}f_{X,Y}(x,y)dy=\int_{\mathbb{R}}\frac{1}{2}\left(\mathbf{1}_{[-1,0]^2}(x,y)+\mathbf{1}_{[0,1]^2}(x,y)\right)dy$$ $$=\frac{1}{2}\int_{\mathbb{R}}\mathbf{1}_{[-1,0]^2}(x,y)dy+\frac{1}{2}\int_{\mathbb{R}}\mathbf{1}_{[0,1]^2}(x,y)dy$$ $$=\frac{1}{2}\mathbf{1}_{[-1,0]}\int_{0}^1dy+\frac{1}{2}\mathbf{1}_{[0,1]}\int_{0}^1dy$$ $$=\frac{1}{2}\left(\mathbf{1}_{[-1,0]}+\mathbf{1}_{[0,1]}\right)$$
For $f_Y$ I will obtain the exact same thing. So what remains to be shown is $$\frac{1}{2}\left(\mathbf{1}_{[-1,0]^2}(x,y)+\mathbf{1}_{[0,1]^2}(x,y)\right)\not=\frac{1}{2}\left(\mathbf{1}_{[-1,0]}+\mathbf{1}_{[0,1]}\right)\cdot\frac{1}{2}\left(\mathbf{1}_{[-1,0]}+\mathbf{1}_{[0,1]}\right)$$
I tried this by integrating over $x$ and $y$ but I will be left with an identity saying $1=1$. But I know they are not independent. Where is my error?
Two things to note about that last equation you wrote. First, observe that the indicator functions on the right should carry variable inputs with them -- i.e. $\mathbf 1_{[-1, 0]}(x)$, etc. Also, note that $$\mathbf 1_{[-1, 0]}(x) + \mathbf 1_{[0, 1]}(x) = \mathbf 1_{[-1, 1]}(x).$$ (This isn't actually quite right because the function should evaluate to $2$ at $x = 0$, but that's just a single point and we can safely ignore it.)
Hence, that last equation should actually be: $$\frac{1}{2}\left(\mathbf{1}_{[-1,0]^2}(x,y)+\mathbf{1}_{[0,1]^2}(x,y)\right)\not=\frac{1}{2}\left(\mathbf{1}_{[-1,1]}(x)\right)\cdot\frac{1}{2}\left(\mathbf{1}_{[-1,1]}(y)\right)$$ and your task is now to find a region on the plane with positive measure for which this equation does not hold. Can you take it from here?