= Solution
Set
$$
f(t)=d(x,\gamma(t)).
$$
The <triangle inequality> makes $f$ <Lipschitz continuous> and hence a <continuous function>. Since $\gamma$ is an isometric embedding,
$$
|t|=d(\gamma(t),\gamma(0))
\leq f(t)+f(0),
$$
so $f(t)\to\infty$ as $|t|\to\infty$. The function is therefore <coercive function>[coercive], and the <extreme value theorem> on a sufficiently large compact interval gives a minimizing parameter.
Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma(t_1)$ and $q=\gamma(t_2)$, and let
$$
R=d(x,p)=d(x,q),
\qquad
L=d(p,q)=|t_1-t_2|.
$$
The restriction of $\gamma$ between the two parameters is a <geodesic> from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the <thin geodesic triangle> condition gives a point $y$ on one of the other two sides with $d(m,y)\leq\delta$. By symmetry suppose $y\in[x,p]$. Then
$$
d(p,y)\geq d(p,m)-d(m,y)\geq L/2-\delta,
$$
so
$$
d(x,m)\leq d(x,y)+\delta
=R-d(p,y)+\delta
\leq R-L/2+2\delta.
$$
But $m$ lies on $\gamma(\mathbb R)$ and $p$ is a closest point, so $R\leq d(x,m)$. Therefore $L\leq4\delta$, which is stronger than the required
$$
|t_1-t_2|\leq6\delta.
$$
This is the <closest point on a geodesic line in a hyperbolic metric space> estimate.
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