Here is Prob. 2, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $x_1 > 1$ and $x_{n+1} \colon= 2 - 1/x_n$ for $n \in \mathbb{N}$. Show that $\left( x_n \right)$ is bounded and monotone. Find the limit.
My Attempt:
We are given that $x_1 > 1$, and $$ x_{n+1} = 2 - \frac{1}{x_n} \tag{0} $$ for all $n \in \mathbb{N}$.
As $x_1 > 1$, so $$0 < \frac{1}{x_1} < 1,$$ which implies that $$ -1 < - \frac{1}{x_1} < 0, $$ and hence $$ 1 < 2 - \frac{1}{x_1} < 2,$$ that is, $$1 < x_2 < 2. \tag{1} $$ Suppose that $k \in \mathbb{N}$ such that $k \geq 2$ and also suppose that $$ 1 < x_k < 2. $$ Then upon taking the reciprocals in the last chain of inequalities, we obtain $$ \frac{1}{2} < \frac{1}{x_k} < 1, $$ and so $$ - \frac{1}{2} > - \frac{1}{x_k} > -1, $$ which implies that $$ 2 - \frac{1}{2} > 2 - \frac{1}{x_k} > 2-1, $$ and hence $$ \frac{3}{2} > 2 - \frac{1}{x_k} > 1,$$ which we can rewrite as $$ 1 < x_{k+1} < \frac{3}{2} < 2. $$ Thus by the Principle of Mathematical Induction, we can conclude that $$ 1 < x_n < 2 \tag{2} $$ for all $n \in \mathbb{N}$ and $n \geq 2$.
Moreover, for each $n \in \mathbb{N}$, we also have $$ x_n - x_{n+1} = x_n - 2 + \frac{1}{x_n} = \left( \sqrt{x_n} - \frac{1}{\sqrt{x_n} } \right)^2 \geq 0, $$ because from (2) it is clear that $x_n$ is a positive real number. Therefore, we can also conclude that $$ x_n \geq x_{n+1} \tag{3} $$ for all $n \in \mathbb{N}$.
From (2) and (3), we find that our sequence $\left( x_n \right)_{n \in \mathbb{N} }$ is a bounded monotonically decreasing sequence; so this sequence is convergent. Let us put $$ x \colon= \lim_{n \to \infty} x_n. $$ Then from (2) we can conclude that $$ 1 \leq x \leq 2, $$ and thence from (0) we obtain $$ x = 2 - \frac{1}{x}, $$ which implies $$ x^2 -2x + 1 = 0, $$ and hence $$x = 1. $$ Therefore, we have $$ \lim_{n \to \infty} x_n = 1. $$
Is my solution correct? If so, then is every step in it correct and lucid enough too? If not, then where is (are) the issue(s)?