I've addressed this as a footnote to the question What is the proper way to explain the twin paradox? but since you are specifically asking about this point I'll post a answer here as well.

You need to know the equation that describes the trajectory of a unifomly accelerating object. The derivation isn't very illuminating so I'll content myself with quoting the result. More more details see Gravitation by Misner, Thorne and Wheeler chapter 6. There is also a summary in Phil Gibbs' excellent article on the relativistic rocket.

Anyhow, suppose you are accelerating with a constant proper acceleration $a$. Note that constant acceleration means the acceleration you feel in your rest frame is constant e.g. you feel a constat $1g$. The acceleration in the rest frame of the inertial observer watching you obviously can't be constant because that observer cannot see you exceed the speed of light. Suppose also that you start at the point $x = c^2/a$ at time $t=0$. In this case your distance as a function of time is given by:

$$ x = \frac{c^2}{a}\sqrt{1 + \left(\frac{at}{c}\right)^2} $$

We take a factor of $at/c$ out of the square root to get:

$$\begin{align} x &= \frac{c^2}{a}\frac{at}{c}\sqrt{1 + \left(\frac{c}{at}\right)^2} \\ &= ct\sqrt{1 + \left(\frac{c}{at}\right)^2} \end{align}$$

Now at time zero we send off a light beam from the origin $x=0$, so the light beam starts off a distance $c^2/a$ behind you. What we're going to show is that as long as you keep your constant acceleration $a$ the light beam can never catch you, and that means there is a coordinate horizon at the distance $c^2/a$ behind you.

To start with we'll simply the last equation. We're interested in what happens at large times $t$, and when $t \gg c/a$ we have $c/at \ll 1$ so we can approximate our equation using the binomial theorem:

$$ x \approx ct\left(1 + \tfrac{1}{2}\left(\frac{c}{at}\right)^2\right) $$

Now the light starts at $x=0$and moves at a constant speed $c$, so the trajectory of the light is given by:

$$x_\text{light}=ct$$

And we get the distance between you and the light beam by subtracting the position of the light from your position:

$$\begin{align} x - x_\text{light} &\approx ct\left(1 + \tfrac{1}{2}\left(\frac{c}{at}\right)^2\right) - ct \\ &\approx \tfrac{1}{2}\left(\frac{c}{at}\right)^2 \end{align}$$

So what we find is that the light beam is always a distance $\tfrac{1}{2}(c/at)^2$ behind you i.e. the light ray can never catch you. This is why there is a horizon for a constantly accelerating observer.

I've addressed this as a footnote to the question What is the proper way to explain the twin paradox? but since you are specifically asking about this point I'll post an answer here as well.

You need to know the equation that describes the trajectory of a uniformly accelerating object. The derivation isn't very illuminating so I'll content myself with quoting the result. For more details see Gravitation by Misner, Thorne and Wheeler chapter 6. There is also a summary in Phil Gibbs' excellent article on the relativistic rocket.

Anyhow, suppose you are accelerating with a constant proper acceleration $a$. Note that constant acceleration means the acceleration you feel in your rest frame is constant e.g. you feel a constant $1g$. The acceleration in the rest frame of the inertial observer watching you obviously can't be constant because that observer cannot see you exceed the speed of light. Suppose also that you start at the point $x = c^2/a$ at time $t=0$. In this case your distance as a function of time is given by:

$$ x = \frac{c^2}{a}\sqrt{1 + \left(\frac{at}{c}\right)^2} $$

We take a factor of $at/c$ out of the square root to get:

$$\begin{align} x &= \frac{c^2}{a}\frac{at}{c}\sqrt{1 + \left(\frac{c}{at}\right)^2} \\ &= ct\sqrt{1 + \left(\frac{c}{at}\right)^2} \end{align}$$

Now at time zero we send off a light beam from the origin $x=0$, so the light beam starts off a distance $c^2/a$ behind you. What we're going to show is that as long as you keep your constant acceleration $a$ the light beam can never catch you, and that means there is a coordinate horizon at the distance $c^2/a$ behind you.

To start with we'll simply reuse the last equation. We're interested in what happens at large times $t$, and when $t \gg c/a$ we have $c/at \ll 1$ so we can approximate our equation using the binomial theorem:

$$ x \approx ct\left(1 + \tfrac{1}{2}\left(\frac{c}{at}\right)^2\right) $$

Now the light starts at $x=0$and moves at a constant speed $c$, so the trajectory of the light is given by:

$$x_\text{light}=ct$$

And we get the distance between you and the light beam by subtracting the position of the light from your position:

$$\begin{align} x - x_\text{light} &\approx ct\left(1 + \tfrac{1}{2}\left(\frac{c}{at}\right)^2\right) - ct \\ &\approx \tfrac{1}{2}\left(\frac{c}{at}\right)^2 \end{align}$$

So what we find is that the light beam is always a distance $\tfrac{1}{2}(c/at)^2$ behind you i.e. the light ray can never catch you. This is why there is a horizon for a constantly accelerating observer.