Difference in frequency spectrum of continuous time signal and discrete time signal?
This is the classical sampling theorem. When you sample a continuous signal $x(t)$, you are basically multiplying it by a sample train $s(t) = \sum_{-\infty}^{+\infty}\delta(t-kT_s)$, the value at consecutive $T_s$ being your samples. In frequency domain, it effect is to convolve $X(\omega)$ and $S(\omega)$

$$S(\omega) = \sum_{-\infty}^{+\infty}\delta(\omega-2\pi k/T_s)$$

So the resulting spectrum would be 
$$
X_s(\omega) = \sum_{-\infty}^{+\infty}X(\omega-2\pi k/T_s)
$$
There are infinite shifted copies of $X(\omega)$. To begin with your original signal was a single-frequency signal which traced the circumference with frequency $f_m$. When you sampled it at $f_s$ rate, you got infinite copies at $f_m \pm kf_s$, where $k$ is any integer in $(-\infty,+\infty)$ hence it referred to as infinite number of lines after sampling (discrete time).