UPDATE: As Zassounotsukushi correctly points out in the comments, my original answer was wrong. I said before that objects move out across our horizon, but they don't. Sorry about that. I hope I've fixed things now.
It's best to avoid phrases like "the Universe is expanding faster than the speed of light." In general relativity, notions like distances and speeds of faraway objects become hard to define precisely, with the result that sentences like that have no clear meaning.
But if we leave that terminological point aside, your questions are perfectly well-posed and physically meaningful.
It's true that we can only see out to a finite distance, due to the Universe's finite age, and that this is an explanation of Olbers's paradox, which is the name for the old puzzle of why the night sky is dark.
As the Universe expands, more objects pop into view, since that "horizon distance" is continually getting bigger, at least in principle. In fact, though, that's a very small effect and would not lead to the night sky becoming brighter in practice.
First, we should note that we wouldn't expect to see stars popping into view as our horizon expands. The reason is that objects right near the edge of the horizon are so far away that we would see them as they were long ago, around the time of the Big Bang. In practice, we can't see all the way back to $t=0$, because the early Universe was opaque, but still, we can see back in time to long before there were discrete objects like stars. When we look back near the horizon, we see a nearly-uniform plasma.
But there's a much more important point. The further away we try to look, the more redshifted the light from a given object is. Even if we could see an object near our horizon, the radiation from it would be shifted to extremely long wavelengths, which also means that it would carry extremely little energy. In practice, this just means that things near our horizon become unobservably faint. In a practical sense, our ability to see faraway objects actually decreases with time: although in principle our horizon grows, the redshift causes any given object to become unobservably faint much faster than the rate at which new stuff is brought in across the horizon.
Lawrence Krauss has written a bunch about this stuff. The details are in this paper, and he has a Scientific American article (paywalled). Dennis Overbye wrote about this stuff in the NY Times a while back too. (If you read the pop stuff, tread carefully. Some of it seems to be saying the incorrect thing I said before, namely that things that are currently inside our horizon move outside of it. The technical article is correct, but the nontechnical ones can be misleading. That's my excuse for messing things up in my original answer, but it's not a very good excuse, because I should've known better.)
The expansion of the universe is due to the expansion of spacetime. There's a good article on this here.
Suppose you take two non-interacting particles, put them some distance apart and make them stationary with respect to each other. If you now watch them for a few billion years you'll see the particles start to accelerate away from each other. This happens because the spacetime between the two particles is stretching i.e. there is more "space" between them.
One way of interpreting the acceleration is to say there is a force between the two particles repelling them. This is a slightly dodgy description because there isn't really a force; it's just expansion of space. Nevertheless, if you tied the two particles together with a rope and watch for a few billion years there would be a tension in the rope so the force is real in this sense.
Anyhow, now we have everything we need to understand why the Earth isn't stretching. The expansion of spacetime creates a stretching force, but this will only have an effect if there is no other force to oppose it. For example you are indeed being stretched by the expansion, but the interatomic forces between the atoms in your body are vastly stronger than the stretching due to expansion, so you remain the same size. Likewise the gravitational force between the Sun and Earth is vastly greater than the stretching force so the Earth's orbit doesn't change.
The stretching force is vanishingly small at small distances, but it gets greater and greater with increasing distance so at some point it wins. Galaxies and indeed galaxy clusters are still too small to be stretched, but at greater sizes than this the stretching wins. That's why galaxy clusters are the largest objects observed in the universe. At greater sizes spacetime expansion wins.
A footnote: if anyone's still interested in this subject, there's a paper Local cosmological effects of order H in the orbital motion of a binary system just out claiming that the effect of the expansion on the Solar System might be measurable.
Best Answer
I have nothing to say about the (possible) infinity of the universe; however, it is the case that infinitely many stars distributed uniformly will create infinite brightness at any point.
Let's throw away all the real physical facts about stars, and make the (incorrect, but good enough for now) assumption that the universe is an infinite volume with some 'star' distributed uniformly throughout it. So there is a uniform 'brightness field' which I'll denote by $\phi$.
Now suppose we are standing at the point $\mathbf 0$. How much brightness are we getting from a small volume $dV$ at a distance $r$ from us? Well, the small volume is emitting light rays at an intensity of $\phi dV$, but as you get further away (say, at a distance $s$) from the volume, the light rays are spread over the surface of a sphere of radius $s$. The surface area of a sphere of radius $s$ is proportional to $s^2$, so the light intensity due to the volume at a point a distance $r$ away from the volume is proportional to $\frac{1}{r^2}$.
(source: ohio-state.edu)
So the volume $dV$ contributes a brightness of $\frac{C}{r^2}dV$ to the total brightness, where $C$ is some constant, and $r$ is the distance of the small volume from $\mathbf 0$. We are now in a position to integrate over a large sphere of radius $R$ to get the brightness due to all the 'stars' at a distance less than $R$ from us:
\begin{align} \textrm{Total brightness at distance less than $R$} &= C\int_{|\mathbf x|<R}\frac1{|\mathbf x|^2}dV\\ &=C\int_0^{2\pi}d\phi\int_0^\pi \sin(\theta)d\theta\int_0^R\frac1{r^2}\times r^2dr\\ &=C\times4\pi\int_0^R dr=4\pi CR \end{align}
So the brightness due to the stars in a sphere of radius $R$ is proportional to $R$. Clearly, if we make $R$ infinitely large, the brightness will become infinitely large as well.
Of course, this ignores several physical realities, such as the expansion of the universe, the geometry of space-time, etc. But that is, at least, the mathematical justification for your friend's claim.