A condition on a non convex subset of $\mathbb{R}^2$ In the euclidean plane $\mathbb{R}^2$ it's possible to state the following theorem:
Theorem: Consider any finite set of $n$ points $\{x_1,\dots,x_n\}\subset \mathbb{R}^2$ and any n-ple of positive real numbers $(r_1,\dots,r_n)$ such that $$\cap_{i=1}^nB_{r_i}(x_i)\neq\emptyset$$ where $B_{r_i}(x_i)$ is the closed ball of center $x_i$ and radius $r_i$. Then for any other set of $n$ points $\{x_1',\dots,x_n'\}\subset \mathbb{R}^2$ such that $$|x'_i-x'_j|\le|x_i-x_j|$$ for every $i,j=1,\dots,n$ it also happens $$\cap_{i=1}^nB_{r_i}(x'_i)\neq\emptyset$$
Question: does this result still hold true if we consider $$X:=\{x\in\mathbb{R}^2 \text{ such that } 0\le\text{arg}(x)\le 7\pi/4 \}$$ instead of $\mathbb{R}^2$? By this I mean that both $\{x_1,\dots,x_n\}$ and $\{x_1',\dots,x_n'\}$ will be in $X$ and the condition $|x'_i-x'_j|\le|x_i-x_j|$ is replaced by $d(x'_i,x_j')\le d(x_i,x_j)$ where $d$ is the induced patch metric by the euclidean metric of $\mathbb{R}^2$ on $X$. Also, the intersections are considered in $X$ and the balls are defined using $d$.
This doesn't seem easy to me because the only proof I know for the theorem in $\mathbb{R}^2$ isn't really easy (it's contained in "A Lipschitz Condition Preserving Extension for a Vector Function" by Valentine and it's for $\mathbb{R}^n$) and doesn't seem to adapt to the case of $X$. As user Moishe Cohen pointed out, I should sketch it.

Lemma 1: (by Alexandrov and Hopf): Let $A_1,\dots,A_n$ be closed sets in $\mathbb{R}^m$ which cover the simplex $T$. Suppose that for each side $a_{i_1},\dots,a_{i_\tau}$ of $T$ it's true $a_{i_1},\dots,a_{i_\tau}\subset A_{i_1}\cup\dots\cup A_{i_\tau}$. Then $A_1\cap \dots\cap A_n\neq \emptyset$.
Lemma 2: Let $\Delta(x_1',\dots,x_n')$ be the euclidean simplex of vertex $x_1',\dots,x_n'$. Under the hypothesis of the theorem $\Delta(x_1',\dots,x_n')$ is covered by the sets $B_{r_i}(x'_i)$
proof of the lemma: suppose $\Delta(x_1',\dots,x_n')$ is not covered by the sets $B_{r_i}(x'_i)$, then we can choose
$$x\in \cap_{i=1}^nB_{r_i}(x_i)\neq\emptyset,\qquad x'\in \Delta(x_1',\dots,x_n')\setminus \cup_{i=1}^nB_{r_i}(x'_i)$$
call $R_i$ the vector from $x$ to $x_i$ and $R_i'$ the vector from $x'$ to $x_i'$. Since $|x'_i-x'_j|\le|x_i-x_j|$ it follows 
$$ |R_i'|>|R_i|, \qquad |R_i'-R_j'|\le|R_i-R_j|$$
for every $i,j=1,\dots,n$. This implies $R_i'\cdot R_j'>R_i\cdot R_j$. Let $(a_1',\dots,a_n')\in \mathbb{R}_+^n$ such that $\sum_{i=1}^na_i'R_i'=0$, then from the previous inequality it follows $\sum_{i=1}^n|a_i'R_i'|^2<0$, which is impossible.
proof of the theorem: since $\cap_{i=1}^nB_{r_i}(x_i)\neq\emptyset$ it follows $B_{r_i}(x_i)\cap B_{r_j}(x_j)\cap \overline{x_ix_j}\neq \emptyset$ and since $|x'_i-x'_j|\le|x_i-x_j|$ it follows $B_{r_i}(x'_i)\cap B_{r_j}(x'_j)\cap \overline{x'_ix'_j}\neq \emptyset$. The result then follows from the two lemmas.

As I said this proof is for the general case of $\mathbb{R}^n$, $n>1$ and is quite complicated. I don't know if for $n=2$ there's a simpler one which would adapt to the case of $X$. I can't understand how to adapt this one because it doesn't seem possible to define $\Delta(x_1',\dots,x_n')$ in $X$ in the first place.
 A: The answer is negative already for $n=3$:
There are points $x_1, x_2, x_3, x_1', x_2', x_3'\in C$ and positive numbers $r_1, r_2, r_3$ for every nonconvex cone $C\subset E^2$ such that:


*

*$d(x_i, x_j)= d(x_i', x_j')$ for all $i, j=1, 2, 3$.

*$$
\{ x\}= \bigcap_{i} \bar{B}(x_i, r_i)\ne \emptyset.
$$

*$$
\bigcap_{i} \bar{B}(x'_i, r_i)= \emptyset.
$$
In my examples $r_1+r_2=d(x_1, x_2)$, $r_i=d(x,x_i), i=1, 2, 3$; thus $x\in x_1x_2$. The point is that for every nonconvex cone $C$, there exists a geodesic triangle $x_1x_2x_3$ and a point $x\in x_1x_2$ such that in the Euclidean comparison triangle $x_1'x_2'x_3'$ and the comparison point $x'\in x'_1x'_2$, we have
$$
d(x'_3, x')> d(x_3, x). 
$$
[To find such examples place $x_1, x_2$ on the different boundary rays of the cone $C$ and place $x_3$ (in the cone interior) so that the geodesic $x_1x_3$ passes through $x=$ the apex of the cone.] 

It follows that 
$$
\bigcap_{i} \bar{B}(x'_i, r_i)= \emptyset,
$$
since 
$$
\bar{B}(x_1',r_1)\cap \bar{B}(x_2',r_2)=x',
$$
but $x'\notin \bar{B}(x_3',r_3)$. Now, the cone $C$ contains arbitrarily large flat disks, so we can place the points $x_i'$ as in your question in such disks, so that they necessarily span flat comparison triangles. 
