TITLE: Connected metric spaces with disjoint open balls
QUESTION [107 upvotes]: Let $X$ be the $S^1$ or a connected subset thereof, endowed with the standard metric. Then every open set $U\subseteq X$ is a disjoint union of open arcs, hence a disjoint union of open balls.
Are there any other metric spaces with this property?
That is: Can you give an example of a connected metric space such that every open set is the union of disjoint open balls and such that it is not homeomorphic to a subspace of $S^1$?

Edit: a few remarks, though most of them leave more questions than they give answers:
If $a,b\in X$, $a\ne b$, then the closed sets $\partial B(a,r)$, $0<r<d(a,b)$ and $\{x\in X\mid d(a,x)=\lambda d(b,x)\}$ for $\lambda>0$ are non-empty because $X$ is connected. Can anything be said about $r\to0$ or $\lambda\to \infty$? For example, does it follow that $X$ is path-connected? 
I wrote "homeomorphic to a subspace of $S^1$" mostly as an abbreviation for "homeomorphic to point, open/half-open/closed bounded interval, or $S^1$-like". The same toppological space may however have the disjoint-ball property with one metric and not have it with another (bounded) metric. For example, in $S^1\subset\mathbb R^2$ with $d\bigl((x,y), (u,v)\bigr) =\sqrt{4(x-u)^2+(y-v)^2}$ the connected open set given by $y>0$ is not a ball: The only candidate is $B\left((0,1),\sqrt5\right)$, but it contains $(0,-1)$.
Does it follow at all that $d$ must be bounded? $X\setminus\{x\}=\bigcup_{i\in I} B(x_i,r_i)$ for some index set $I$, $x_i\in X$, $r_i>0$. For $\epsilon>0$, the ball $B(x,\epsilon)$ must intersect every $B(x_i,r_i)$ because $X$ is connected, hence $d(x,x_i)=r_i$ for all $i\in I$ and $d(x,y)\le2\sup\{r_i\mid i\in I\}$. Thus if $X\setminus\{x\}$ has only finitely many components for some $x$, then $d$ is bounded. But could  $X\setminus\{x\}$ have infinitely many components for all $x$?  Resolved after a comment from celtschk: $X$ itself is a ball $B(x_0,r)$, hence $d(x,y)\le d(x,x_0)+d(x_0,y)<2r$ for $x,y\in X$.
Assume a point $x_0$ looks like a finite branch, that is there is $n\in\mathbb N$, $n\ge3$ and a continuous map $h:\{1,\ldots,n\}\times[0,\epsilon)\to B(x_0,\epsilon)$ such that $d(x_0,h(i,t))=t$ and $h|_{\{1,\ldots,n\}\times(0,\epsilon)}\to B(x_0,\epsilon)\setminus\{x_0\}$ is a homeomorphism.
Given $\mathbf r=(r_1, \ldots, r_n)$ with $0<r_i<\epsilon$, the set $U_{\mathbf r}=h\left(\bigcup_{i=1}^n\{i\}\times[0,r_i) \right)$ is open and connected, hence a single open ball $B(\hat x,r)$. Let $x_i=h(i,r_i)$. By continuity of $h$, we conclude $d(\hat x,x_i)=r$ and the $r_i$ can be recovered from $\hat x$ and $r$ as $r_i=\inf\{t\mid d(\hat x, h(i,t))\ge r\}=\sup\{t\mid d(\hat x,h(i,t))<r\}$. One checks that this $\phi_i\colon(\hat x, r)\mapsto r_i$ is continuous where defined. This gives a contiunuous and surjective map $(i,s,t)\mapsto (\phi_i(h(i,s),t))_i$ from a subset of the twodimensional $\{1,\ldots,n\}\times (0,\epsilon)^2$ to the $n$-dimensional $(0,\epsilon)^n$. Unless this is some space-filling monster (can it be?), we conclude that $X$ cannot have branch-points. (This is ugly - can it be made prettier? Or branch-point be defined friendlier?)

Edit: Meanwhile I am confident that every connected length space with the disjoint open ball property is one of the known spaces (i.e. homeomorphic to a connected subspace of  $S^1$). So, how far is a connected metric space from being a length space? Could the ideas be transferred or do they give hints for counterexamples?
In what follows, let $(X,d)$ be a connected length space with the disjoint open ball property.
We can define the branch degree (or is there a standard name for this?) $\beta(x)$ for $x\in X$ as the (possibly infinite) number of connected components of $X\setminus\{x\}$.
Lemma 1: For $x\in X$, we have $\beta(x)\le2$.
Proof: Assume $\beta(x)\ge 3$, i.e. $X\setminus \{x\}$ has connected components $U_i$, $i\in I$ and wlog. $\{1,2,3\}\subseteq I$.
For $i\in\{1,2,3\}$ select a point $x_i\in U_i$ and let $\rho=\min\{d(x,x_1),d(x,x_2),d(x,x_3)\}$. Then for $r<\rho$ and $i\in\{1,2,3\}$ we have that $U_i\cap B(x,r)$ is connected because an (approximately) geodesic path to $x$ cannot leave the connected component and stays within the ball. Also, we can find a point $\in U_i$ at distance $r$ from $x$. 
The set $$U:=(U_1 \cap B(x,\rho))\cup(U_2 \cap B(x,\frac12\rho))\cup B(x,\frac13\rho)\cup\bigcup_{i\in I\setminus\{1,2,3\}}U_i$$ is open and connected, hence $U=B(y,R)$ for some $y\in X$, $R>0$. As  paths between points in different $U_i$ must pass through $x$, we conclude that $R=d(x,y)+\rho$ if $y\notin U_1$, $R=d(x,y)+\frac12\rho$ if $y\notin U_2$ and $R=d(x,y)+\frac13\rho$ if $y\notin U_3$. Since $y$ is in at most one of $U_1, U_2, U_3$, we arrive at a contradiction.$_\blacksquare$
I think the following should be possible to prove:
$Lemma 2:$ If $a,b,c$ are three distinct points $\in X$, then $X\setminus\{a,b,c\}$ is not connected.
Proof: ???
Since I'm not yet sure about a proof of lemma 2, the rest is left in handwaving stage:
Assume there exists $x\in X$ with $\beta(x)=0$. Then $X$ is just a point and we are done.
Assume there exists $x\in X$  with $\beta(x)=2$. Write $X\setminus\{x\}=U_1\cup U_2$ and definie $f:X\to\mathbb R$ by $$f(y)= \begin{cases}d(x,y)&y\in U_1\\-d(x,y)& y\in U_2\end{cases}$$
I claim that $f$ is injective and in fact it should be possible to show this with Lemma 2 or some similar result.
Then we are left with the case that $\beta(x)=1$ for all $x$. Then for any such point either $X\setminus\{x\}$ should have a point $y$ with $\beta(y)=2$ hence $X\setminus\{x\}$ "is" an interval and we conclude that $x$ is one of its endpoints (or possibly "is both" endpoints, making an $S^1$).
Or otherwise at least $X\setminus\{x,y\}$ must have a point $z$ with $\beta(z)=2$ (by lemma 2) and hence $X\setminus\{x,y\}$ "is" an interval and $x,y$ are its endpoints.

REPLY [2 votes]: I do not know what happens for general metric spaces, here is a proof which covers the case of length metric spaces. 
Definition. Call a metric space $X=(X,d)$ a B-space if every open subset of $X$ equals the disjoint union of some open metric balls in $X$. 
Note that each length metric space $(X,d)$ satisfies the following two properties:
a. $X$ is locally path-connected. (Since open balls in $X$ are path-connected.)
b. Every closed metric ball $\bar B(a,r)=\{x\in X: d(a,x)\le r\}$ equals the closure of the corresponding open ball 
$B(a,r)=\{x\in X: d(a,x)< r\}$. 
I will use the notation $S(a,r)$ for the metric sphere $S(a,r)=\{x\in X: d(a,x)=r\}$. 
Theorem 1. Suppose that $X$ is a  $B$-space which consists of more than 1 point and which satisfies (a) and (b). Then $X$, is a topological 1-dimensional manifold (possibly with boundary). In other words, $X$ is homeomorphic to a connected subset of $S^1$. 
Proof. It suffices to consider the case when $X$ is connected. Since $X$ is a Hausdorff path-connected space, it is also arc-connected, i.e. any two points in $X$ belong to an arc, i.e. a subset homeomorphic to a closed interval, see 
S. Willard, General Topology, Addison-Wesley 1970. Theorem 31.2.  (Compare the discussion here.) 
Given an arc $\alpha\subset X$ which is the image of a homeomorphism $f: [0,1]\to \alpha$, let $\alpha^\circ$ denote 
$f((0,1))$, the corresponding open arc. 
Since $X$ is locally connected, each $B(a,r)$ contains a unique maximal open connected subset containing $a$, I will denote this subset $B_c(a,r)$. 
Lemma 1. Each open  arc $\alpha^\circ$ is an open subset of $X$. 
Proof. To prove this, consider for each $r>0$ the open $r$-neighborhood $U_r(\alpha)$ in $X$:
$$
U(\alpha,r)=\bigcup_{a\in \alpha} B(a,r). 
$$
This neighborhood will contain an open connected sub-neighborhood of $\alpha$: 
$$
U_c(\alpha,r)= \bigcup_{a\in \alpha} B_c(a,r). 
$$
Since $X$ is a B-space, there are points $a_r\in X$ and radii $R=R(r)$ such that 
$$
U_c(\alpha,r)= B(a_r, R(r)).
$$
For each $r$ there exists $x_r\in \alpha$ such that $d(x_r, a_r)<r$. By compactness of $\alpha$, there is a sequence $r_i\to 0+$ such that 
$$
x_{r_i}\to a\in \alpha 
$$
and $R(r_i)\to R(0)$. Since $\alpha$ is not a singleton, $R(0) >0$. 
Clearly, $a_{r_i}\to a$, 
$$
\bigcap_{r_i} U_c(\alpha,r)= \alpha 
$$
and 
$$
\alpha= \bar{B}(a, R(0)). 
$$
We conclude from this that $\alpha\cap B(a,R(0))$ is an open subset of $X$. From this, we see that the subset of $X$ where $X$ is a 1-dimensional manifold is dense (and clearly open) in $X$. So far, we did not use the Property (b). 
Claim. I claim that $\alpha^\circ \cap S(c, R(0))=\emptyset$. 
Proof. If $\alpha^\circ \cap S(a, R(0))\ne \emptyset$, the arc $\alpha$ contains a  subarc $\beta$ connecting  points of $S(a, R(0))$ and not containing the center $a$. Let $b\in \beta$ be a point with minimal distance from $a$; by our assumption that $a\notin \beta$, $\rho=d(b,a) > 0$. Since $\alpha\cap B(a, R(0))$ is a 1-dimensional manifold of $B(a, R(0))$, the point $b$ cannot be the limit of any sequence $b_i\in B(a, R(0))$ such that $(d(a,b_i))_{i\in {\mathbb N}}$ converges to $d(a,b)$ from the left. This contradicts the Property (b) (the point $b\in S(a,\rho)$ does not belong to the closure of $B(a, \rho)$.)  qed 
The claim implies that $\alpha^\circ = \alpha \cap B(a, R(0))$, i.e. $\alpha^\circ$ is open in $X$. This concludes the proof of Lemma 1. qed 
Lemma 1 implies that $X$ is a 1-dimensional manifold near each point contained in an open arc. Suppose, therefore, that $x\in X$ is a point which does not belong to any open arc. Since $X$ is not a singleton, $x$ is the end-point of some nondegenerate arc $\alpha\subset X$ connecting points $x$ and $y$ in $X$.
Lemma 2.  I claim that $\alpha-\{y\}$ is a neighborhood of $x$. 
Proof. If not, there is a sequence $x_n\to x$, $x_n\notin \alpha$. Since $X$ is locally arc-connected, there exist arcs $\beta_n$ connecting $x_n$ to $x$ and disjoint from $y$ for all sufficiently large $n\ge n_0$. Lemma 1 implies that $\beta_n\cap \alpha= \{x\}$ for all $n\ge n_0$. Therefore, the concatenation $\gamma_n$ of $\beta_n$ and $\alpha$ is an arc in $X$ (for $n\ge n_0$) such that $x\in \gamma_n^\circ$. A contradiction. qed 
Combining the Lemmata 1 and 2 we see that $X$ is a 1-dimensional manifold (possibly with boundary). qed