Suppose we are working in $\mathbb{R}$ with the standard topology.
Suppose further the following hypothesis:
Let $U$ be an open set. For every $x\in U$ there exists a unique $\varepsilon >0$ such that the open ball $B(x,\varepsilon)\subset U$.
Let $\mathcal{C} = \{ (U_{\alpha},x_{\alpha}) \}_{\alpha\in\Lambda}$ be a possibly uncountable collection of pointed open sets. That is to say, for each $\alpha$ we are given a point $x_\alpha\in U_\alpha$.
I want to define a function $f: \mathcal{C}\to \mathbb{R}$ such that $f((U_\alpha,x_\alpha))\in U_\alpha\setminus \{ x_{\alpha} \} $. In other words, I want to define a choice function that picks out a point of $U_\alpha$ different from the point $x_\alpha$ already existing in the data.
Let $(a_\alpha,b_\alpha)$ be the unique open ball such that $x_\alpha\in (a_\alpha,b_\alpha) \subset U_\alpha$. Then I can define $f$ as follows: $$ f((U_\alpha,x_\alpha)) = \frac{1}{2}\left(\frac{1}{2}(a_\alpha+b_\alpha) + b_\alpha\right) $$
Did I define $f$ without having to invoke the axiom of choice?
EDIT: Apparently the above hypothesis is demonstrably false, so I will revise it to the following hypothesis, which is less clear that it is false:
Let $U$ be an open set. For every $x\in U$ there exists a unique open ball $B(x,\varepsilon_x)\subset U$.
As written, what you have proceeds from a false hypothesis. But guessing what you are trying to do, let's note the following:
Proposition. Let $U$ be an open subset of $\mathbb{R}$, and let $x\in U$. If $U\neq\mathbb{R}$, then $\epsilon_{x,U} = \sup\{\epsilon\gt 0\mid B(x,\epsilon)\subseteq U\}$ exists, is positive, and $B(x,\epsilon_{x,U})\subseteq U$.
Proof. Call the set of the epsilons described in the statement $E$.
Since $U$ is open, the $E$ nonempty. Note that if $\epsilon\gt 0$ is in $E$, and $0\lt\epsilon'\lt \epsilon$, then $\epsilon'\in E$. Since $U\neq\mathbb{R}$, there exists $y\in\mathbb{R}$, $y\notin U$. If $d=|x-y|\gt 0$, then $B(x,2d)$ is not contained in $U$, since $y\in B(x,2d)$. Thus, the set is bounded above, and hence it has a supremum $\epsilon_{x,U}$ which is positive.
Now let $z\in B(x,\epsilon_{x,U})$; then $|x-z|\lt\epsilon_{x,U}$, so there exists $\epsilon\in E$ such that $|x-z|\lt\epsilon\leq \epsilon_{x,U}$; hence $z\in B(x,\epsilon)$, so $z\in U$ because $\epsilon\in E$. Thus, $B(x,\epsilon_{x,U}) \subseteq U$, as desired. $\Box$
Now, given an open set $U$ that is nonempty and not all of $\mathbb{R}$, we define a function $f_U\colon U\to (0,\infty)$ by $f_U(x) = \epsilon_{x,U}$. This does not require choice: we are using the order of $\mathbb{R}$ to define $f_U$.
You can now use these functions to define your function, essentially as you give; you need say what to do if $U=\mathbb{R}$, but that is straightforward.