Why is the Fourier transform of a Dirac comb a Dirac comb? This doesn't make sense to me, because the Heisenberg inequality states that $\Delta t\Delta \omega$ ~ 1.

Therefore when you have something perfectly localized in time, you get something completely distributed in frequency. Hence the basic relationship $\mathfrak{F}\{\delta(t)\} = 1$ where $\mathfrak{F}$ is the Fourier transform operator.

But for the Dirac comb, applying the Fourier transform, you receive another Dirac comb. Intuitively, you should also get another line.

Why does this intuition fail?
I believe that the fallacy is to believe that a Dirac comb is localized in time. It isn't because it is a periodic function and as such it can only have frequency components at multiples of its fundamental frequency, i.e. at discrete frequency points. It can't have a continuous spectrum, otherwise it wouldn't be periodic in time. Just like any other periodic function, a Dirac comb can be represented by a Fourier series, i.e. as an infinite sum of complex exponentials. Each complex exponential corresponds to a Dirac impulse in the frequency domain at a different frequency. Summing these Dirac impulses gives a Dirac comb in the frequency domain.