Suppose that $(X, \rho)$ is a metric space, $|X| > 1$. Is it possible to prove that $X$ is an open set without assuming the axiom of choice?
As I understand it, the challenge is to find a way to assign an open ball to each point of the space (assign a radius, basically). The way I found is to select some point $y \in X$ and assign to each point $x \in X\setminus\!\{y\}$ the ball $B_{2\rho(x,y)}(x)$. Then it is easy to show that the union of these balls is $X$, and so $X$ is open.
Is there another way to prove that $X$ is open without relying on the axiom of choice? Or am I wrong and the axiom of choice is not required for the above proof?