Okay, since the comments aren't doing it, I'm gonna have a go at explaining time dilation in this answer.
When <PERSON> introduced the special theory of relativity in his 1905 paper "On the Electrodynamics of Moving Bodies," he outlined two postulates. The first, that the laws of electrodynamics and optics are valid for all frames of reference for which the equations of mechanics hold good, and the second, that the speed of light is constant for all observers, regardless of motion. This is important, it means that regardless of your speed relative to the light, you always observe it travelling at $c$ (299,792,458 m/s) (well, until you get into General Relativity, where warped space can make light appear to slow down, but that's beside the point).
What you need to realize is that there is no absolute frame of reference. When you are in an inertial reference frame (no acceleration), there is no way of telling if you're moving or if the things around you are moving. In fact, from your point of view, you're stationary.
Now, let's get into what the question is about. Imagine two mirrors spaced at a distance $L$ apart (as pictured). Now imagine a light beam is projected from the bottom mirror at time $t_0$ in the direction of the top mirror; this beam is then reflected off the top mirror at time $t_1$ before returning to the bottom mirror at time $t_2$. The time between $t_0$ and $t_2$ can be expressed as
$$\Delta t=\frac{2L}{c}$$
Now imagine you see both mirrors start moving uniformly to the right. For someone moving with the mirrors, they will be stationary in their reference frame, so the situation will play out exactly as described above. 562
However, for you, the light takes a longer path. After leaving the first mirror at time $t_0$, the light traverses the new distance $D$ and hits the top mirror at time $t_1^\prime$ before reflecting back into the bottom mirror at time $t_2^\prime$. The time between $t_0$ and $t_2^\prime$ can be expressed as
$$\Delta t^\prime=\frac{2D}{c}=\frac{2\sqrt{L^2+(\frac{1}{2}v\Delta t^\prime)^2}}{c}$$
Now, you may be thinking, "wouldn't this be the same as if you bounced a ball off the ceiling of a moving train?" The answer is no. That ball has the velocity of the train added on to it and so to a stationary observer it moves with a higher velocity than what the person bouncing the ball sees. That is not the case with light. Remember that second postulate I mentioned? What that means is that both the observer moving with the mirrors and the observer stationary with regards to the moving mirrors record the light moving at the same speed. Think about that. From the point of view of the stationary observer the light traces out a longer path and takes longer to hit the top mirror: there is more time between "ticks".
This is what <PERSON> realized. Moving clocks run slower from the perspective of a stationary observer.
Also, you know all that stuff I said about no absolute inertial reference frame? That means that the observer moving with the mirrors also sees the other observer's clock run slow; that's because to them, they're stationary and the other observer is moving.
I hope that cleared things up. 562