Suppose we are working on a bounded interval $[a,b]$ with Sturm-Liouville operator $L$ given by $$ Lf = \frac{1}{w(x)}\left[-\frac{d}{dx}\left(p(x)\frac{df}{dx}\right)+q(x)f\right]. $$
How can I prove that the eigenvalues of the operator can be ordered as an increasing sequence such that : $\lambda_0 < \lambda_1 < \lambda_2... < \lambda_n < ... \to + \infty$
On
The problem needs to be a regular problem on $[a,b]$; otherwise what you say may not be true at all. If the problem is regular, and if you have endpoint conditions $$ \cos\alpha f(a)+\sin\alpha f'(a) = 0 \\ \cos\beta f(b)+\sin\beta f'(b) = 0, $$ then you basically get what you want. One way to argue is by looking at the classical solutions of the Sturm-Liouville equation $Lf=\lambda f$ with endpoint conditions $$ f(a) = -\sin\alpha,\;\;\; f'(a)=\cos\alpha. $$ Then $f(x,\lambda)$ and $f'(x,\lambda)$ will depend holomorphically on $\lambda$, which means that the solutions where $\cos\beta f(b,\lambda)+\sin\beta f'(b,\lambda)=0$ will correspond to the zeroes of a holomorphic function of $\lambda$. Then some simple estimates of $\langle Lf,f\rangle$ will show that this form is bounded below, and, therefore, the eigenvalues are bounded below. Knowing that the eigenvalues are the zeros of a power series in $\lambda$ that is not identically $0$ tells you that the set of eigenvalues do not cluster and, therefore, can be ordered.
What's your knowledge in basic Functional Analysis? If enough, note that it is consequence of the spectral theorem for compact operators (having in mind that eigenvalues for Sturm-Liouville-type operators are not defined in the classical way, but inversed).
See https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Spectral_theorem.