I saw in a lecture (see after 00:55:00) by Leonard Susskind, he write Polyakov action, saying that the string is supposed to have divided into several pieces with mass (he says that it is the non-relativistic analogue mass) reducing proportionately and disappear eventually as the limit approaches infinity and each is connected with the neighbouring ones with springs.
My question is, if one assumes all strings to have a finite mass, then how does massless particles arise in the theory?
Susskind corrects himself at 1:00:08 where he says that the "mass density" coefficient in front of the $\dot{X}^2$ term is a non-relativistic analogue; it is not actually a mass-density of the relativistic string. Instead it is proportional to the string tension $T_0$.
This is related to the fact that the non-relativistic string actually displays a Lorentz symmetry. However, the characteristic speed $c$ is then not the speed of light but the speed of sound, cf. e.g. this Phys.SE post.
To see how masses enter string theory, see e.g. this Phys.SE post.
To see how to go take the continuum limit in the discrete non-relativistic string (which is the main theme of Susskind's lecture), see e.g. this Phys.SE post.
Here is a somewhat hand waving explanation that I think is appropriate given the nature of the question. The equations for the positions of the little masses along the string is essentially a wave equation and can be expanded into modes where both the wave number and frequency are proportional to an integer $n$ indexing the modes. This is the story even for harmonics on an ordinary classical string. The proportionality constant between the frequency and the mode index $n$ is where the mass per length and the tension comes in (due to the equation of state of the relativistic string these are actually the same thing)
We want to quantize the string and each mode is like a distinct harmonic oscillator. The lowest energy state of string is when each one of these modes is in its ground state, but in quantum mechanics the ground state energy of the nth mode $E_{0,n}$ is not simply zero it is $$E_{0,n}=\frac{\hbar \omega_n}{2}\propto n.$$ The ground state energy of the mode is proportional to $n$ since the frequency $\omega_n$ is. So if we sum up all the zero point energies over all the modes we get something like $$\sum_n E_{0,n}\propto\sum_n n=-\frac{1}{12}.$$ This is the part you are not going to like, but the idea is any sensible way of regularizing this divergent sum over zero point energies leads to a negative answer. If you learn a lot more formalism than is present in the Susskind lectures you can phrase this calculation in such a way that the sum over $n$ is never encountered in the first place.
This means that actually the lowest energy (or mass) state of the string is actually negative. It is the so called tachyon state that really reflects an instability in the bosonic string theory. If we look at the next excited states of the string some extra positive energy is added by the excited harmonic oscillator modes, and it turns out that extra energy is exactly right to bring it to zero mass.
