Bounded Geodesics in a Quadrangle of Fixed Length. The following question is a slightly weaker version of the question presented here: Geodesic Quadrangle in a Hyperbolic Space. However, this bound suffices for the statement mentioned in the original question.
Suppose $[p,q]\cup[q,r]\cup[r,s]\cup[s,p]$ is a geodesic quadrangle in a $\delta$-hyperbolic space $X$, such that $d(p,q)=d(s,r)$. Then for any pair of points $x\in [p,q], y\in [r,s]$ with $d(p,x)=d(s,y)$ we have
$$d(x,y)≤3\max\{d(p,s),d(q,r)\}+10δ.$$
 A: We break the proof into three cases.
Case 1: $d(s,x)\leq\max\{d(s,p),d(r,q)\}+5\delta$.
In this case $x$ and $y$ are close enough to the "pq-side" of the quadrangle that following the path along this side serves as a sufficient bound.
Formally, the path $[x,s]*[s,p]*[p,y]$ has $$l([x,s]*[s,p]*[p,y])\leq l([x,s])+l([s,p])+l([p,y])$$ $$\leq (\max\{d(s,p),d(r,q)\}+5\delta)+d(s,p)+(\max\{d(s,p),d(r,q)\}+5\delta) \leq 3\max\{d(p,s),d(q,r)\}+10δ.$$
Case 2: $d(x,r)\leq\max\{d(s,p),d(r,q)\}+5\delta$.
This is the same as Case 1, but on the other side of the quadrangle.
Case 3: $d(s,x)>\max\{d(s,p),d(r,q)\}+5\delta$ and $d(x,r)>\max\{d(s,p),d(r,q)\}+5\delta$.
First, by the $\delta$-slimness of quadrangles, there exists $w\in[s,p]\cup[p,q]\cup[q,r]$ with $d(x,w)\leq2\delta$. Note that $w\not\in[s,p]$. If it were, we would have $d(s,x)\leq d(s,w)+d(w,x)\leq d(s,p)+2\delta$, contradicting our assumption on $d(s,x)$. A similar, symmetric argument shows that $w\not\in[q,r]$. Therefore, $w\in[p,q]$.
Our goal is to now show that $w$ is close to $y$, as then the geodesics from $x$ to $w$, then to $y$ will be small. We will do this by showing that a construction where $w$ is far away from $y$ is impossible.
Let $w\in[p,y]$.$^{[1]}$ For the sake of contradiction, let us assume that $d(w,y)>d(p,s)+5\delta$. Then $$d(w,p)=d(p,y)-d(y,w)< d(p,y)-(d(p,s)+5\delta).$$
Which is to say,
$$d(w,s)+5\delta\leq d(w,p)+d(p,s)+5\delta<d(p,y).$$
However this quickly shows a contradiction to the definition of $w$:
$$d(w,s)+5\delta<d(p,y)=d(s,x)\leq d(s,w)+d(w,x)\quad\Rightarrow\quad 5\delta<d(x,w).$$
We therefore conclude that $d(w,y)\leq d(s,p)+5\delta$, which gives $d(x,y)\leq d(x,w)+d(w,y)\leq 2\delta+d(s,p)+5\delta$. $\square$
[1] If we instead assume that $w\in[y,q]$, then comparing $d(x,w)$ to $d(r,q)$ produces a similar argument.
