I'll start with a more careful version of what the article is trying to say, and then I'll give two derivations, one using a straightforward calculation and one using geometric intuition, to explain why the statement is true.
A more careful description
The article's description is carelessly worded, because an observer can't "see" a light ray reach anything far away unless/until light returns to the observer. Here's a more careful version of what it's trying to say:
Consider an inertial observer in AdS spacetime. Consider a moving object that leaves the observer, turns around after reaching some finite distance,$^\dagger$ and then returns to the observer, such that the time between the two meetings is equal to $\Delta \tau$ according to the observer's own internal clock. Between the two meetings, the object moves slower than the speed of light. The faster it moves, the farther away it can be when it turns around while still keeping the time between meetings fixed at the same value $\Delta \tau$ on the observer's own clock. This is true both in flat spacetime and in AdS spacetime. In flat spacetime, the distance to the turnaround event remains finite even if the object is moving at the speed of light, but in AdS spacetime, the distance to the turnaround event becomes infinite if the object moves at the speed of light. That's a more accurate version of what the article was trying to say. The article doesn't explain how this is possible, but it is possible, and I'll show this using two different methods.
$^\dagger$ By "distance," I mean the proper distance measured along a spacelike radial line connecting the turnaround event to the observer's worldline halfway between the two meetings (see below).
Derivation using straightforward calculation
The metric of AdS spacetime can be written like this:
$$
d\tau^2 = A(r)dt^2 - \frac{dr^2}{A(r)} - \text{angular part}
\tag{1}
$$
with
$$
A(r) = 1+\frac{r^2}{L^2}
\tag{2}
$$
where $L$ fixes the scale of curvature. Equation (1) gives the elapsed proper time $d\tau$ along any infinitesimal segment of a worldline, where $dt$ and $dr$ are the infinitesimal coordinate intervals. I'm not writing the angular part because we only need to consider radial worldlines. Or we can just work in 1+1 dimensional AdS spacetime, which is sufficient for the question, and then the "angular part" is absent.
In the scenario described above, suppose that the "observer" sits at $r=0$ for all $t$, and suppose that the "object" is described by the worldline
$$
r = L\tan\left[\left(\frac{\pi}{2}- \frac{|t|}{L} \right)\omega\right]
\tag{3}
$$
where $0<\omega<1$ is a constant that controls the object's speed. The two meetings occur at $t=\pm(\pi /2)L$, and the observer's proper time between meetings is $\Delta\tau = \pi L$, independent of $\omega$. We can substitute (3) into (1) to get
$$
d\tau^2 = \frac{1-\omega^2}{\cos^2\big((\pi/2-|t|/L)\omega\big)}dt^2.
\tag{4}
$$
If $\omega<1$, then this worldline is timelike ($d\tau^2>0$), so it represents the motion of an object moving less than the speed of light (which equals $1$ in the units I'm using here). As $\omega\to 1$, the worldline approaches the speed of light ($d\tau\to 0$), and the value of $r$ at the turn-around time $t=0$ approaches $r\to \infty$, as claimed.
Derivation using geometric intuition
I'll start by describing a geometric construct, and then I'll explain how it answers the question.
Here's the geometric construct. Consider a 3d space with coordinates $(t_1,t_2,x)$. Choose a positive real number $L>0$, and consider the 2d submanifold defined by
$$
t_1^2+t_2^2 = L^2+x^2.
\tag{5}
$$
This is a hyperboloid. The intersection of the hyperboloid (5) with the plane $x=0$ is a circle. If $a\geq b>0$, then the intersection of the hyperboloid (5) with the plane $ax+bt_1=0$ is an ellipse, except in the limiting case $a=b$ where it becomes a pair of straight lines (because the ellipse becomes infinitely elongated). Independently of $a$ and $b$, the ellipse intersects the $x=0$ circle at the two points $(t_1,t_2,x) = (0,\pm L,0)$. This remains true even in the limiting case $a=b$, where the ellipse becomes infinitely elongated, leaving a pair of (infinitely long) straight lines separated from each other by (and parallel to) the plane $t_2=0$.
To relate this to the question, choose the metric on the hyperboloid (5) to be such that a segment $(dt_1,\,dt_2,\,dx)$ is timelike, lightlike, or spacelike according to whether the quantity
$$
dt_1^2 + dt_2^2 -dx^2
\tag{6}
$$
is positive, zero, or negative, respectively. With such a metric, the surface of the hyperboloid is a 1+1 dimensional spacetime, and we can use the angle in the ambient $t_1$-$t_2$ plane as a timelike coordinate in the 1+1 dimensional spacetime. This construct includes 1+1 dimensional AdS spacetime as a special case. (The AdS spacetime is curled up with circumference $2\pi L$ in the time dimension, but that doesn't matter for the question, because we only need to consider the $t_1\geq 0$ part of the hyperboloid.) The half-circle defined by $x=0$ and $t_1\geq 0$ represents the worldline of the "observer" in the 1+1 dimensional AdS spacetime. This part of the observer's worldline has finite proper time (arc-length $\pi L$ in the 3d ambient space). The half-ellipse defined by $ax+bt_1=0$ and $t_1\geq 0$ represents the worldline of an object that leaves the observer at the event $(t_1,t_2,x) = (0,-L,0)$ and returns to the observer at the event $(t_1,t_2,x) = (0, L,0)$. If $a>b$, then the object's worldline is timelike, because (6) is positive everywhere on the ellipse. In the limiting case $a=b$, the object's worldline is lightlike because (6) is zero on the "ellipse." Since the ellipse is infinitely elongated in this case, it represents a light ray that leaves the observer, travels to infinity, and then returns to the observer. Regardless of $a,b$, the observer's proper time between the leave-event and the return-event has the same finite value, because these two events are given by $(t_1,t_2,x) = (0,\pm L,0)$ regardless of the values of $a$ and $b$.
