Is it possible to prove that the metric space is an open set without choice? 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?
 A: You can pick the radius of the open ball centred at $x$ however you like, and the union of all the balls will be $X$. For example, you could just let all the radii be 1.
A: There is no choice issue here at all, and no need to pick a particular radius, since in the definition of open set, there is no obligation for us to assign to each point a unique radius. Rather, we can assign many radii to each point, and this will also verify opennes. We just need to know that every point has at least one ball around it contained in the set; perhaps there are many balls around that point contained in the set (and indeed, of course any smaller ball also works). So let us simply take all balls around any point of any possible radius to see that the whole space is open.
A: $X$ is an open set because every topology (=collection of open sets) on $X$ must contain $X$, see the definition. Nothing metric here.
A: As an aside, note that you don't need Choice to validate the openness of an arbitrary open subset $U\subseteq X$.  That is, we can assign an open ball around each $u\in U$ without Choice: let $\delta_u = \frac12 \inf\{\rho(u,x) : x\notin U\}$
and set $F_u = B_{\delta_u}(u)$.
