Local geodesics in a hyperbolic space are uniform quasi geodesics

First let $\delta >0$ and fix $[s,t]\subseteq I$. Put $L=t-s$, $n=\lfloor L\mathrm{/}(3\delta )\rfloor$, $\sigma =L-3n\delta$, and $x_i=q(s+3i\delta )$ for $0\le i\le n$. Consecutive triples lie on an isometrically parametrized subpath of length $6\delta$. For $0\le i<n$ put $H_i=\{z:d(z,x_{i+1})\le d(z,x_i)\}$ and $B_i=\{z:d(z,x_{i+1})=d(z,x_i)\}$. For $i<n-1$, F2 gives $H_{i+1}\subset \{z:d(z,x_{i+1})<d(z,x_i)\}\subset H_i$ and $\mathrm{dist}(B_i,B_{i+1})\ge \delta$.

Now suppose $\delta =0$. A geodesic segment is closed: if $p$ has distance zero from the image of $\gamma :[0,\mathrm{\ell }]\to X$, set $r=d(\gamma (0),p)$. Arbitrarily close image points $\gamma (u)$ satisfy $\mathrm{\mid }u-r\mathrm{\mid }\le d(\gamma (u),p)$, so $r\in [0,\mathrm{\ell }]$ and $d(\gamma (r),p)=0$. A zero-slim degenerate triangle consisting of two segments between the same endpoints and a constant side consequently forces their images to agree; radial parameters then agree too. For any triangle $(a,b,c)$ let $\alpha =(b\mathrm{\mid }c{)}_a$. The points of $[a,b],[a,c]$ at radius $\alpha$ coincide: apply F3 twice, through $b$ and $c$, to see their product at $a$ is at least $\alpha$ and hence their distance zero. Call the common point $m$. The equality $d(b,c)=d(a,b)+d(a,c)-2\alpha =d(b,m)+d(m,c)$ shows that the concatenated tails form the unique geodesic $[b,c]$. Uniqueness prevents any further common tail. Thus every triangle is a tripod.

Suppose $n\ge 2$. A geodesic $\gamma :[0,d(x_0,x_n)]\to X$ from $x_0$ to $x_n$ starts outside every $H_i$ and ends inside every $H_i$ by nesting. For each $i$ its first entry time $t_i$ into $H_i$ exists and lies in $B_i$: the inverse image of $H_i$ is nonempty and closed, so its infimum belongs to it, and continuity of the distance difference forces equality at first entry. This closed-infimum assertion follows by taking real parameters approaching the infimum; continuity passes the nonpositive inequality to their limit. Nesting gives $t_i\le t_{i+1}$. F2 now gives $t_0\ge 3\delta \mathrm{/}2$, $t_{i+1}-t_i\ge \delta$, and $d(x_0,x_n)-t_{n-1}\ge 3\delta \mathrm{/}2$. Adding yields $d(x_0,x_n)\ge (n+2)\delta \ge n\delta$. For $n=0,1$, the bound $d(x_0,x_n)\ge n\delta$ follows directly from locality.

The final subarc has length $\sigma <3\delta$. Thus $d(q(s),q(t))\ge n\delta -\sigma =L\mathrm{/}3-4\sigma \mathrm{/}3\ge L\mathrm{/}3-4\delta$. Arc length gives $d(q(s),q(t))\le L\le 3L+4\delta$. Since $s,t$ were arbitrary, this is the required quasi-geodesic inequality. Increasing the locality radius preserves its $6\delta$ hypothesis.

Subdivide any compact parameter interval into finitely many equal positive pieces of length less than $k\mathrm{/}2$; a zero-length interval already is geodesic. Each piece is geodesic. Suppose the path through the first $j$ pieces is geodesic and append the next piece at $b$. In the tripod formed by the old starting point, $b$, and the new endpoint, failure of their concatenation to be geodesic would mean both segments from $b$ initially follow the same positive-length leg. Choose $h>0$ smaller than that leg and both adjacent mesh lengths. The points at parameter distances $h$ before and after the join would coincide, although their parameter distance is $2h<k$. This contradicts locality. Finite induction proves the compact restriction geodesic, hence the whole interval map is geodesic. Only finitely many auxiliary segments have been chosen for each restriction; AC is not used.