"Standard" proof that open disks in $\mathbb{R}^2$ are connected? Homework for a complex analysis course asks me to prove as homework that open disks are connected. I do know a way to do this: open disks are convex, and an old exercise in Rudin's "Principles of Mathematical Analysis" guarantees that convex sets are connected. However, I suspect I'm missing the point a bit here.
Is there a quicker, more "first-principles" proof?
 A: A nice conceptual proof would be to notice that an open disk is homeomorphic to $\mathbb R^2$ - for instance, the map
$$f(x,y)=\left(\frac{x}{1-x^2-y^2},\frac{y}{1-x^2-y^2}\right)$$
is such a homeomorphism. Then, choose your favorite proof that $\mathbb R^2$ is connected (or, better, that $\mathbb R^2=\mathbb R\times \mathbb R$ is the product of two connected sets, and hence is connected).
A: You just have to show, that for an open disk $D$ and $x,y\in D$, the connecting line
$$\{\lambda x+(1-\lambda)y\ |\ \lambda\in [0,1]\}$$
is a subset of $D$ since this implies that $D$ is path connected. 
This can be done using the triangle inequality: Let $z$ be the center of the disk, and $R$ it's radius. From $x,y\in D$ it follows, that
$$\|x-z\|<R,\ \|y-z\|< R$$
and therefore
$$\|\lambda x+(1-\lambda)y-z\| =\|\lambda x+(1-\lambda)y-(\lambda+1-\lambda)z\|\leq \lambda\|x-z\|+(1-\lambda)\|y-z\| < R$$
A: Since $x \mapsto \|x-x_0\|$ is a convex function, then set $B(x_0,r) = \{ x \mid \|x-x_0\| < r \}$ is convex. Convexity implies path connected (in fact, straight line connected) which implies connected.
Since you are in $\mathbb{R}^2$, you could note that
$B(x_0,r) = \phi([0,r) \times [0, 2 \pi))$,
where $\phi(r,\theta) = x_0+(r \cos \theta, r \sin \theta)$. Since $[0,r) \times [0, 2 \pi)$ is connected and $\phi$ is continuous, it follows that
$B(x_0,r)$ is connected.
