Connected length space with disjoint open ball property This question deals with a special case of this question, which has not yet been satisfactorily solved. If you have any ideas about that general case, feel free to answer there and I'll be happy to close/delete this question.
Let $(X,d)$ be a metric space such that every open set is the disjoint union of open balls. Also assume that $(X,d)$ is a length space, i.e. for all $x,y\in X$ and $L>d(x,y)$ there exists a weakly contracting curve $\gamma\colon [0,L]\to X$ from $x$ to $y$ (that is, we have $\gamma(0)=x$, $\gamma(L)=y$ and $d(\gamma(t_1),\gamma(t_2))\le |t_1-t_2|$ for $t_1,t_2\in[0,L]$).
Conjecture: $(X,d)$ is homeomorphic to a connected subspace of $S^1$ (that is, one of $\{0\}$, $(0,1)$, $[0,1)$, $[0,1]$, $S^1$.
Note that the more general question was about connected spaces and that length spaces are connected. I have a strong feeling that the case of length spaces is a lot simpler.
 A: Here are my thoughts so far:
In what follows, curves $\gamma$ are always weak contractions. Lemma 0 and 1 are general remarks about length spaces.
Lemma 0. If $\gamma\colon[0,L]\to X$ is a weak contraction with $x=\gamma(0)\ne \gamma(L)=y$, then we may assume wlog. that $\gamma$ is injective.
Proof:
Let $S$ be the set of all finite unions of intervals of the form 
$$ A=\bigcup_{k=1}^n[a_k,b_k]$$
with $a_1=0$, $b_n=L$, $a_k<b_k<a_{k+1}$, $\gamma(b_k)=\gamma(a_{k+1})$.
For $A\in S$ we have $\mu(A)=\sum (b_k-a_k)\ge d(x,y)$ and can define the weak contraction 
$$\gamma_A\colon[0,L-\mu(A)]\to X , t\mapsto \gamma(\mu(A\cap[0,t])).$$
Let $(A_n)_n$ be a sequence in $S$ such that $\mu(A_n)$ converges to $m:=\sup\{\,\mu(A)\mid A\in S\,\}$.  One checks that $(\gamma_{A_n}|_{[0,L-m]})_n$ converges pointwise to a weak contraction $[0,L-m]\to X$ that is injective.$_\square$
Lemma 1. Every open ball is (path-)connected.
Proof:  Indeed, If $y\in B_r(x)$, then there is a curve $\gamma$ from $x$ to $y$ that stays within $B_r(x)$, e.g. we can take $L=\frac{r+d(x,y)}2$.$_\square$
Lemma 2. If $\gamma$ is a curve from $x$ to $y\ne x$ with $L<d(x,y)+\epsilon$, then for every $t\in(3\epsilon,L-3\epsilon)$ there is a neighbourhood of $\gamma(t)$ that is homeomorphic to an open interval.
Proof: Let 
$$\tag1 U=\bigcup_{0<t<L}B_{\min\{\epsilon,t,L-t\}}(\gamma(t)).$$
As union of open sets, $U$ is open. Moreover, the single balls are connected by lemma 1 and are overlapping. This makes $U$ connected.
For some $t_0\in(3\epsilon,L-3\epsilon)$, consider $$\tag2U'=U\setminus\{z\}\quad\text{with }z=\gamma(t_0).$$
By assumption, $U'$ can be written as disjoint union of open balls. 
If we assume that $U'$ is still connected, this union is in fact just a single ball $B_r(\xi)$.
Because $U'$ overlaps every ball around $x$ (or $y$ or $z$), we see that $d(\xi,x)=d(\xi,y)=d(\xi,z)=r$, hence $$\tag32r=d(\xi,x)+d(\xi,y)\ge d(x,y)>L-\epsilon.$$
Since there is a $\tau\in[0,L]$ with $d(z,\gamma(\tau))<\epsilon$, we have $$\tag4r=d(\xi,x)<\tau+\epsilon$$ and $$\tag5r=d(\xi,y)<(L-\tau)+\epsilon,$$ hence $r<\frac L2+\epsilon$. Also, $$\tag6r=d(\xi,z)<\epsilon+|\tau-t_0|.$$
If $t_0\le \tau$, this implies $r<\epsilon+\tau-t_0$ and together with 
(5) and (3)
$$ (L-t_0)+2\epsilon>2r>L-\epsilon.$$
If on the other hand $t_0>\tau$, then (6), (4) and (3) impliy
$$ t_0+2\epsilon>2r>L-\epsilon.$$
In both cases we obtain a contradiciton to $t_0\in(3\epsilon,L-3\epsilon)$. 
Therefore, $U'$ is not connected.
Now for a point $\eta\in U'$, the following cases are possible:


*

*$\eta\in B_\epsilon(z)$. Then also $\eta\in B_\epsilon(z')$ for some $z'=\gamma(t')$ with $t'\approx t_0$. Therefore there is a path within $U'$ from $\eta$ to $\gamma(t')$ and from there along $\gamma$ to either $x$ or $y$.

*$\eta\notin B_\epsilon(z)$. Then of course $\eta\in B_{\min\{\epsilon,t',L-t'\}}(\gamma(t'))$ for some $t'\ne t_0$ and again $\eta$ is path-connected within $U'$ to $x$ or $y$.
We conclude that $U'$ has exactly two connected components, one containing $x$ and one containing $y$.
Let $\delta =\min\{\epsilon,t_0-3\epsilon,L-3\epsilon-t_0\}>0$.
Assume that $B_\delta(z)$ contains a point $\eta$ that is not in the image of $\gamma$.
For $t_1$ sufficiently close to $t_0$ (and with $t_0<t_1<t_0+\delta$, say), the point $\eta$ is also in $B_\epsilon(\gamma(t_1))$.
By the same argument as above, in the set $U''=U\setminus\{\gamma(\frac{t_0+t_1}2)\}$ the points $x$ and $y$ are in different connected components. But in fact they are connected via paths $x\to \gamma(t_0)\to\eta\to\gamma(t_1)\to y$.
Therefore $B_\delta(z)$ contains only points in $\gamma([0,L])$ and is homeomorphic image of a connected subset of $[0,L]$.$_\square$
Lemma 3. If $x\ne y$ then there is a curve $\gamma\colon[0,L]$ from $x$ to $y$ such that $\gamma|_{(0,L)}$ is a homeomorphism with an open subset of $X$.
Proof: 
Using lemma 2 with $\epsilon<\frac16d(x,y)$ we find a point $z=\gamma(\frac L2)$ that has a neighbourhod looking like an interval. Any (injective) curve from $z$ to $x$ taking the "wrong" direction has a length of at least $\frac L2+\epsilon$, hence this does not happen for all sufficiently short  curves from $z$ to $x$. We conclude that a sequence of curves from $z$ to $x$ where $L\to d(z,x)$ converges pointwise to a curve $[0,d(z,x)]\to X$ such that the restriction to $[0,d(z,x))$ is a homeomorphism. If we do the same with $y$ instead of $x$ and combine the results, the claim of the lemma follows.$_\square$
