I've read that $C = \{(x,y) \mid x>0, y<0 \}$ is an open set in $\mathbb{R}^n$, but I think that I proved that it is not and I don't know what is wrong with my reasonings.
If we take for instance the point $c = (1,-1)$, then for every $r>0$, the open ball $B(c,r)$ will "come out" of x,y plane and thus it will include points that are not in $C$, therefore C is not open...
This is probably wrong, since I've read a convincing proof that says the opposite..
So what do you think is wrong? Thanks!!
It's an open set in $\mathbb{R}^2$, and that is also the only context this makes sense, as $C$ is a set of pairs. So it's not a member of a higher $\mathbb{R}^n$ at all, and so it makes no sense to talk about being open there.
The analogous set $\{(x_1,\ldots,x_n) \in \mathbb{R}^n: x_1 > 0, x_2 < 0\}$ is also open in $\mathbb{R}^n$ as it equals $\pi_1^{-1}[(0,\infty)] \cap \pi_2^{-1}[(-\infty,0)]$, where $\pi_1, \pi_2$ are the projections onto the first two coordinates.
The word "ball" in the context of metric spaces does not necessarily mean a three dimensional ball. Of course, if you put a three-dimensional ball around a point in a plane, you get "outside" the plane. But here, in your case, all you need to do is to check that there is a two-dimensional "ball" around $(1,-1)$ that lies entirely in the set $C$, where a two-dimensional "ball" is just a circle. So take, for instance, a circle of radius $1/2$ and it will work.
The other answers are fine for your particular problem, but let's define an open ball $B_{r}(x)$ in a metric space $(M,d)$ where $r>0$ is the radius of the ball and $x\in M$ its center.
$$B_{r}(x)=\left\{y\in M\,\vert\,d(x,y)<r\right\}$$
so that you can see, when $(M,d)=(\mathbb{R}^{2},d_{E})$ where $d_{E}$ denotes the usual euclidean distance, that $$B_{r}(x_{0},y_{0})=\left\{(x,y)\in \mathbb{R}^{2}\,\vert\,\sqrt{(x_{0}-x)^{2}+(y-y_{0})^{2}}<r\right\}$$ which is indeed an usual disk.
