I am really confused about the proof of this theorem: For any continued fraction, $$q_n\alpha - p_n = \frac{(-1)^n}{\alpha_{n+1}q_n + q_{n-1}}$$
I got the base induction case for $n=0$ but I can't figure out how to finish the induction step. Thanks for your help!
Well then. I'm not sure how much you know about continued fractions, so if need be, refer to here where I set up some standard properties of continued fractions, albeit modulo different notation.
In particular, we have the equalities
$$p_{n-1}q_n-p_nq_{n-1}=(-1)^n$$ $$[a_0; a_1, \ldots, a_{m-1}, x]=\frac{xp_{m-1}+p_{m-2}}{xq_{m-1}+q_{m-2}}$$ $$\alpha=[a_0; a_1, \ldots, a_n, \alpha_{n+1}]$$
Thus
\begin{align} q_n\alpha-p_n=q_n[a_0; a_1, \ldots, a_n, \alpha_{n+1}]-p_n& =q_n\frac{\alpha_{n+1}p_n+p_{n-1}}{\alpha_{n+1}q_n+q_{n-1}}-p_n\\ &=\frac{p_{n-1}q_n-p_nq_{n-1}}{\alpha_{n+1}q_n+q_{n-1}}\\ &=\frac{(-1)^n}{\alpha_{n+1}q_n+q_{n-1}} \end{align}
For a proof of the equalities I use (which are standard well known properties in continued fractions), refer to my linked answer above, albeit with different notation.
If that doesn't prove useful, you might want to consider picking up a book which covers continued fractions and learn from there.