The proof is as follows:
line 1: Suppose that $x_1 = 1$ and $$x_{n+1} = \frac{x_n +4}{x_n +1}$$
line 2: Suppose that $x_n$ converges to $s$; then the left-hand side of this equation converges to $s$ (by the shift rule), and the right-hand side of this equation converges to $\frac{s+4}{s+1}$ (by the quotient and sum rules, provided that $s \neq -1$). So, if $x_n \rightarrow s$, we would have $$s = \frac{s+4}{s+1} \Rightarrow s(s+1)=s+4 \Rightarrow s^2 = 4 \Rightarrow s = \pm 2$$
line 3: Since the terms are positive, we guess that $x_n \rightarrow 2$; we now prove this. From before, we have $$x_{n+1}-2 = \frac{x_n +4}{x_n +1} - 2 = \frac{x_n +4-2x_n -2}{x_n +1} = \frac{2-x_n}{x_n +1} = -\frac{x_n -2}{x_n +1}$$
line 4: We can prove that $x_n \ge 1$ for all $n$ by induction: it's true for $n = 1$ as $x_1 = 1$. If it's true for $n = k$, then $x_k \ge 1$, and so we have $$x_{k+1} = \frac{x_k +4}{x_k +1} \ge 1$$
line 5: So we have $x_n \ge 1$ for all $n \in \mathbb{N}$. It now follows that $$\vert{x_{n+1} -2}\vert = \frac{\vert{x_n -2}\vert}{\vert{x_n +1}\vert} \le \frac{1}{2} \vert{x_n -2}\vert$$
The line below is the one I don't understand. More specifically, how do they generalize $\frac{1}{2}\vert{x_n -2}\vert$ into $2^{-n}\vert{x_1 -2}\vert$?
line 6: so, we can say that $$\vert{x_{n+1} -2}\vert \le 2^{-n}\vert{x_1 -2}\vert = 2^{-n} = (\frac{1}{2})^n $$
line 7: Therefore $x_n -2 \rightarrow 0$ as $n \rightarrow \infty$, so $x_n \rightarrow 2$. $\square$
Appendix:
The shift rule :
Where $n, N \in \mathbb{N}$, it can be said that $\sum_{n=1}^\infty{a_n}$ converges to a number if and only if $\sum_{n=1}^\infty{a_{n+N}}$ converges.
The quotient rule and sum rule:
Suppose that $a_n \rightarrow a$ and $b_n \rightarrow b$. Then we can say that $\frac{a_n}{b_n} \rightarrow \frac{a}{b}$ and that $a_n + b_n \rightarrow a + b$, respectively.