**edit**

**Theorem** Let $X$ be a metric space containing a homeomorphic copy of the interval $(0, 1)$. Then the regular open sets of $X$ do not admit a uniformly continuous choice function.

I'll show just the case $X = (-2, 2)$ (the interval) and skip the epsilon-delta details and the fact there could be stuff around the embedded path, since the details of this are very similar to the original (see below).

(Note that a choice function admitting an $f$-metric choice function just means uniform continuity from $(S, d_H|_{S \times S})$ to $X$ with $f$ the modulus of continuity.)

For $n \in \mathbb{Z}$ define
$$ U_n = (\arctan(n)/\frac{\pi}{2} - \epsilon_n, \arctan(n)/\frac{\pi}{2} + \epsilon_n) $$
where $\epsilon_n$ are sufficiently small so that these sets are disjoint. So we have "order type $\zeta$ many" open intervals side by side inside $(-1,1) \subset X$. Each $U_n$ is a regular open set in $(-1,1) \subset X$, and $U_n \cap U_m = \emptyset$ if $n \neq m$. The union of all these, $U = \bigcup_n U_n$, is also easily seen to be regular open.

Now suppose $g$ is a choice function for regular opens. Then $g(U) \in U_n$ for some $n \in \mathbb{Z}$. Slide $U_L = \bigcup_{m \leq n} U_m$ continuously to the left side of $X$, join it to a single component and morph it into the interval $V_L = (-5/3,-4/3)$. Slide then $U_R = \bigcup_{m > n} U_m$ to the right side, join it to a single component and morph it to $V_R = (4/3,5/3)$. The choice must follow along, i.e.
$$ g(U) \in U_L \implies g(V_L \cup V_R) \in V_L. $$

But if we define
$U_L' = \bigcup_{m < n} U_m$ and $U_R' = \bigcup_{m \geq n} U_m$, and do the exact same with these sets, we get
$$ g(U) \in U_R' \implies g(V_L \cup V_R) \in V_R. $$

That is the contradiction that squares up the proof.

**original**

OP has suggested that I write an answer based on my comment. Here's one possible statement you get from that idea, quick write-up, I'll fix later if I screwed up the epsilons.

Let $X$ be a metric space and $S \subset \mathcal{P}(X)$ a set of sets in $X$. Let $f : \mathbb{R}_+ \to \mathbb{R}_+$ be a function. A function $g : S \to X$ is an **$f$-metric choice function for $S$** if $g(A) \in A$ for all $A \in S$, and $d(g(A), g(B)) \leq f(d_H(A, B))$ for all $A, B \in S$. We say $S$ then **admits an $f$-metric choice function**.

**Theorem** Let $f : \mathbb{R}_+ \to \mathbb{R}_+$ satisfy $\lim_{x \to 0} f(x) = 0$ and let $X$ be a metric space containing a homeomorphic copy of $S^1$. Then the regular open sets of $X$ do not admit an $f$-metric choice function.

Proof. Let $h : S^1 \to X$ be the embedding of $S^1$ into $X$, and let $\epsilon > 0$ be such that opposite points on $S^1$ map at least distance $\epsilon$ apart form each other in the map $h$. Let $0 < \delta < \epsilon/10$ be such that $f(x) < \epsilon/10$ for $x < 3\delta$.

Identify $S^1$ as $\mathbb{R}/\mathbb{Z}$. To each $a \in S^1$ associate the set
$$ k(a) = k_1(a) \cup k_2(a) $$
where
$$ k_1(a) = \overline{B_{\delta}(h(a))}^\circ $$
and
$$ k_2(a) = \overline{B_{\delta}(h(a+1/2))}^\circ \subset X. $$
If $\delta > 0$ is small enough, $k(a)$ is regular open for all $a$. (The interior of the closure of an open set is regular open, so $k_i(a)$ is. The union of two regular opens may not be regular open in general, but since $\delta < \epsilon/10$ this happens.) Again because $\delta < \epsilon / 10$, the sets $k_1(a)$ and $k_2(a)$ are disjoint.

Suppose we had a choice function $g$ for the regular opens that is $f$-continuous. W.l.o.g. we may assume $g(k(a)) \in k_1(a)$ for some $a \in S^1$. Then by picking small enough increments, it is easy to see that in fact $g(k(a)) \in k_1(a)$ for **all** $a \in S^1$.

(Here's some algebra to show that in case it's not obvious: If the distance between $h(a)$ and $h(a')$ is at most $\delta$, then the distance between $k_1(a)$ and $k_2(a')$ is at least $\epsilon - 3\delta > \epsilon/10$, and
$$ d_H(k(a), k(a')) \leq \max(d_H(k_1(a), k_1(a')), d_H(k_2(a), k_2(a'))) \leq 3\delta, $$
so $g(k(a')) \in k_1(a')$ whenever $g(k(a)) \in k_1(a)$ and $|a'-a|$ is small enough.)

But now we have a contradiction since
$$ g(k(a)) \in k_1(a) $$
and
$$ g(k(a)) = g(k(a+1/2)) \in k_1(a+1/2) = k_2(a) $$
and $k_1(a) \cap k_2(a) = \emptyset$. Square.

4more comments