Uniform incentre of collection of quasi convex subspaces in hyperbolic spaces I'm reading a paper of Wise on cubulations and the following fact is used:
Let $H$ be a quasi-convex subgroup of a $\delta$-hyperbolic group and let $H_i, i\in I$ be a finite family of translates of $H$ such that they are pairwise at bounded distance $D$. Then there is a constant $C$ and a point  $x$ such that $d(x,H_i)<C$. The constant $C$ does not depend on $|I|$ but only on $\delta, D$ and the constant of quasi-convexity.
My question is: Does that remain true without group assumption? Namely:
Let $X$ be a $\delta$-hyperbolic space and let $H_i, i\in I$ be a finite collection of quasi-convex subspaces of $X$, with the same constant $Q$ of quasi-convexity for all of them. Suppose that there is $D$ such that $d(H_i,H_j)<D$ for all $i,j$. Is it true that there is $C$ depending only on $\delta, Q, D$ and a point $x$ such that $d(x,H_i)<C$ for all $i$?
Remark: such an $x$ plays the role of the incentre of the family $H_i$. It is easy to show by induction that for any $|I|$ such a $C$ exists, the point is exactly the independence of $C$ (the inradius) from $|I|$. 
 A: Yes, your statement is true. A proof can be adapted from Theorem 6.1 in this paper. Since the argument is elementary, I write a proof below.
Proposition: Let $X$ be a $\delta$-hyperbolic space and $\{C_i : i \in I \}$ a finite collection of $Q$-quasiconvex subspaces such that $C_i^{+D} \cap C_j^{+D} \neq \emptyset$ for every $i,j \in I$. Then 
$$\bigcap\limits_{i \in I} C_i^{+6(15\delta+\max(D,Q))} \neq \emptyset.$$
The definition of $\delta$-hyperbolicity I use is the same as the one used in Bowditch's course. I pick there three classical lemmas:
Lemma 1: [Bowditch, Lemma 6.5] Let $X$ be a $\delta$-hyperbolic space. For every geodesic triangle $[x,y] \cup [y,z] \cup [z,x]$, one has $[x,y] \subset \left( [y,z] \cup [z,x] \right)^{+6 \delta}$.
Lemma 2: [Bowditch, Lemma 6.2] Let $X$ be a $\delta$-hyperbolic space. For every points $x,y,z \in X$ and every geodesic $[y,z]$, one has $d(x,[y,z]) \leq (y,z)_x +4\delta$.
Lemma 3: [Bowditch, Lemma 6.4] Let $X$ be a $\delta$-hyperbolic space and $x,y \in X$ two points. The Hausdorff distance between any two $(1,C)$-quasigeodesics between $x$ and $y$ is at most $\frac{3}{2}C+12\delta$.
Now, the key lemma is the following:
Lemma 4: Let $X$ be a $\delta$-hyperbolic space, $C_1,C_2 \subset X$ two $Q$-quasiconvex subspaces satisfying $C_1^{+D} \cap C_2^{+D} \neq \emptyset$, and $z \in X$ a point. For $i=1,2$, fix a point $z_i \in C_i$ satisfying $d(z,z_i) \leq d(z,C_i)+\delta$, and suppose that $d(z,z_1) \geq d(z,z_2)$. Then $z_1 \in C_2^{+6(15\delta+\max(D,Q))}$.
Proof. Fix a point $u \in C_1^{+D} \cap C_2^{+D}$, and for $i=1,2$ a point $u_i \in C_i$ satisfying $d(u,u_i) \leq D$.
Fact 1: $d(z,C_i) \leq d(z,[u,z_i]) +6\delta +\max(D,Q)$. 
Indeed, fix a point $z' \in [u,z_i]$ such that $d(z,[z_1,u]) \geq d(z,z')- \epsilon$ for some small $\epsilon$. According to Lemma 1, there exists some $z''\in [u_i,z_i] \cup [u_i,u]$ such that $d(z',z'') \leq 6\delta$. If $z'' \in [u,u_i]$, then
$$d(z,C_i) \leq d(z,z')+d(z',z'')+d(z'',C_i) \leq d(z,[u,z_i]) + 6\delta +D+\epsilon;$$
and if $z'' \in [u_i,z_i]$, then
$$d(z,C_i) \leq d(z,z')+d(z',z'')+d(z'',C_i) \leq d(z,[u,z_i])+6 \delta+Q + \epsilon.$$
Taking $\epsilon \to 0$, our fact is proved. 
Fact 2: $[u,z_i]\cup [z_i,z]$ is a $(1,2(11\delta+\max(D,Q)))$-quasigeodesic between $u$ and $z$. 
We deduce from Lemma 1 that
$$\begin{array}{lcl} (u,z_i)_z & \geq & d(z,[u,z_i])- 4 \delta \geq d(z,C_i) - 10\delta - \max(D,Q) \\ \\ & \geq & d(z,z_i) -11\delta- \max(D,Q) \end{array}$$
By definition, $(u,z_i)_z= \frac{1}{2} \left( d(z,u)+d(z,z_i)-d(u,z_i) \right)$, so we get
$$d(z,u) \geq d(u,z_i)+d(z_i,z) - 2(11\delta+\max(D,Q)),$$
which proves our second fact.
Now we are ready to conclude the proof of our lemma. It follows from Lemma 3 that there exists some $p \in [z,z_2] \cup [z_2,u]$ satisfying $d(z_1,p) \leq 3(15\delta+\max(D,Q))$. If $p \in [z,z_2]$, then
$$d(p,z_2) = d(z,z_2)-d(z,p) \leq d(z,z_2)-d(z,z_1)+d(z_1,p) \leq d(z_1,p)$$
hence
$$d(z_1,C_2) \leq d(z_1,p)+d(p,z_2) \leq 2d(z_1,p) \leq 6(15\delta+\max(D,Q)).$$
If $p \in [z_2,u]$, then
$$d(z_1,C_2) \leq d(z_1,p)+d(p,C_2) \leq 51 \delta +4 \max(D,Q)$$
since $[z_2,u]$ is contained into the $(6\delta+ \max(D,Q))$-neighborhood of $C_2$. This concludes the proof of the lemma. $\square$
Proof of the proposition. For every $i \in I$, fix a point $z_i \in C_i$ such that $d(z,z_i) \leq d(z,C_i)+\delta$. Pick up some $j \in I$ such that $d(z,z_j) \geq d(z,z_k)$ for every $k \in I$. Applying Lemma 4 provides
$$z_j \in \bigcap\limits_{k \in I} C_k^{6(15 \delta + \max(D,Q))},$$
which concludes the proof. $\square$
Remark: If you want, you can replace the assumption that $I$ is finite with the assumption that the $C_i$'s all intersect a given bounded set.
