a) Suppose the contrary, that is $f$ has another fixed point $x_1 \in (0, \infty) \setminus \{x_0\}$. Then $\lim_{n \to \infty} (f^n(x_0) - f^n(x_1)) = x_0 - x_1 \ne 0$, a contradiction. In this case, the continuity of $f$ (intermediate value property) implies $f(x) < x$, for all $x \in (0, x_0)$ or $f(x) > x$, for all $x \in (0, x_0)$.

By induction, the first case implies $0 < f^{n+1}(x) < f^n(x) < x$, for any $n \in \mathbb{N}^*$ and $x \in (0, x_0)$. It follows that the sequence $a_n = (f^n(x))_{n \ge 1}$ is convergent; denote by $a$ its limit. If $a > 0$, then from $a_{n+1} = f(a_n)$ we get $a = f(a)$, a contradiction. So $a = 0$, which implies $\lim_{n \to \infty} (f^n(x_0) - f^n(x)) = x_0 \ne 0$, for all $x \in (0, x_0)$, contradiction. In conclusion $f(x) > x$, for any $x \in (0, x_0)$.

Analogously, $f(x) > x$, for all $x \in (x_0, \infty)$ or $f(x) < x$ for all $x \in (x_0, \infty)$. In the first case we deduce $f^{n+1}(x) > f^n(x) > x$, for any $n \in \mathbb{N}^*$, implying

$$\lim_{n \to \infty} f^n(x) = \infty$$
and then $\lim_{n \to \infty} (f^n(x) - f^n(x_0)) = \infty$, for all $x \in (x_0, \infty)$, a contradiction. So, $f(x) < x$, for any $x \in (x_0, \infty)$.

b) First solution. Consider $x, y \in (0, \infty)$. We may suppose $f(x) < f(y)$. Denote $x_n = f^n(x)$, $n \in \mathbb{N}^*$ and $y_n = f^n(y)$, $n \in \mathbb{N}^*$. We have $2 \le x_n < y_n$, $n > 1$, because $f$ is increasing on $[1, \infty)$. We shall prove inductively that $y_n < y_1 + 2\sqrt{n}$, $n \in \mathbb{N}^*$. This is obvious for $n=1$. Supposing $y_n < y_1 + 2\sqrt{n}$, for some $n$, by the monotonicity of $f$, we get: $y_{n+1} = f(y_n) < f(y_1 + 2\sqrt{n}) = y_1 + 2\sqrt{n} + \frac{1}{y_1 + 2\sqrt{n}} < y_1 + 2\sqrt{n} + \frac{1}{2\sqrt{n}} < y_1 + 2\sqrt{n+1}$.

It follows $2 \le x_n < y_n < y_1 + 2\sqrt{n} < 3\sqrt{n}$, for $n \in \mathbb{N}$, $n > y_1^2$. Using the well-known inequality $1 - x < e^{-x}$, for $x \in \mathbb{R}$, we deduce: $0 < y_{n+1} - x_{n+1} = f(y_n) - f(x_n) = (y_n - x_n) \left(1 - \frac{1}{x_n y_n}\right) < (y_n - x_n) \left(1 - \frac{1}{9n}\right) < (y_n - x_n) e^{-\frac{1}{9n}}$, for $n \ge p = [y_1^2] + 1$. We get $0 < y_n - x_n < (y_p - x_p)e^{-\frac{1}{9} \sum_{k=p}^{n-1} \frac{1}{k}}$, for all $n > p$.

Because $\lim_{n \to \infty} \sum_{k=p}^{n-1} \frac{1}{k} = \infty$, we infer $\lim_{n \to \infty} e^{-\frac{1}{9} \sum_{k=p}^{n-1} \frac{1}{k}} = 0$, which in turn gives $\lim_{n \to \infty} (f^n(y) - f^n(x)) = 0$. So $f$ is contractive without fixed points.

Second solution. Using the same notations as before, we shall show that $\lim_{n \to \infty} (x_n - \sqrt{2n}) = \lim_{n \to \infty} (y_n - \sqrt{2n}) = 0$, which will imply $\lim_{n \to \infty} (x_n - y_n) = 0$, that is the conclusion. As $x_{n+1} = x_n + 1/x_n$, $n \ge 1$, we have $\lim_{n \to \infty} x_n = \infty$. By the Stolz–Cesàro lemma, we deduce

$$\begin{align*} \lim_{n \to \infty} (x_n - \sqrt{2n}) &= \lim_{n \to \infty} \frac{x_n^2 - 2n}{x_n + \sqrt{2n}} = \lim_{n \to \infty} \frac{x_{n+1}^2 - x_n^2 - 2}{x_{n+1} - x_n + \sqrt{2n+2} - \sqrt{2n}} = \\ &= \lim_{n \to \infty} \frac{1/x_n^2}{1/x_n + 2/(\sqrt{2n+2} + \sqrt{2n})} = \\ &= \lim_{n \to \infty} \frac{1/x_n}{1 + 2x_n/(\sqrt{2n+2} + \sqrt{2n})} = 0. \end{align*}$$

The last equality is a consequence of $0 \le \frac{1/x_n}{1 + 2x_n/(\sqrt{2n+2} + \sqrt{2n})} \le 1/x_n$ and $\lim_{n \to \infty} 1/x_n = 0$.