Is it possible to define a function for every point in its domain such that every point maps to its family of open balls without the axiom of choice?

Question

Suppose I want to define a function $f:X\to \mathcal{F}$ where $(X,d)$ is a metric space and $\mathcal{F}$ is a set of a family of all open balls for every $x\in X$, more formally ¹:

$\displaystyle \mathcal{F}:=\{ Z \in \mathcal{P}\mathcal{P}(X) : (\exists x\in X)(\forall Y\in \mathcal{P}(X)) \left[ Y\in Z \leftrightarrow (\exists \epsilon>0) (Y=B(x,\epsilon)\right] \}$

Now if I can show $(\forall x\in X)(\exists ! Z\in\mathcal{F})(\forall Y\in Z)\left[x\in Y\right]$, can I then use this formula to define $f(x):=Z$ for that unique $Z$ without appealing to the axiom of choice?

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Footnote.

1. Apologies for the readability as I eschewed using the axiom of choice by ensuring that the set formed did not use it. The set $T:=\{\{B(x,\epsilon) : \epsilon>0\}_{x} \in \mathcal{P}\mathcal{P}(X) : x\in X \}$ is exactly the set $\mathcal{F}$ above, but I'm not sure if I needed the axiom of choice to form $T$.