Given the recursively defined sequence $(x_n)_{n\in\mathbb{N}}\subseteq\mathbb{R}$:
$x_1=1$
$x_{n+1}=1+\frac{1}{x_n}$
How can I prove that $x_{2n-1}\leq x_{2n+1}$ and $x_{2n}\geq x_{2n+2}$ for all natural numbers without $0$ ?
That is, that the sub-sequence of all odd indices is growing and the sub-sequence of all even indices is falling?
Writing out a few elements of your sequence, we have $1, 2, 3/2, 5/3, 8/5, \dots$. This is clearly the ratio of successive Fibonacci numbers, as you probably already noticed based on your tags. To prove this we'll use induction. We have already shown our base step via direct evaluation in the first sentence, so for our inductive step, assume $x_n = \frac{f_{n+1}}{f_n}$ where $f_n$ is the $n$th Fibonacci number. Then
$$x_{n+1} = 1 + \frac{1}{x_n} = 1 + \frac{1}{\frac{f_{n+1}}{f_n}} = 1 + \frac{f_n}{f_{n+1}} = \frac{f_n + f_{n+1}}{f_{n+1}} = \frac{f_{n+2}}{f_{n+1}}$$
which proves the claim. Now we can rewrite $x_{2n-1} \leq x_{2n+1}$ as
$$\frac{f_{2n}}{f_{2n-1}} \leq \frac{f_{2n+2}}{f_{2n+1}} \implies f_{2n+2}f_{2n-1} - f_{2n}f_{2n+1} \geq 0$$
Expanding out $f_{2n+2}$ and $f_{2n+1}$, we have
$$(f_{2n+1} + f_{2n})f_{2n-1} - f_{2n}(f_{2n} + f_{2n-1}) = f_{2n+1}f_{2n-1}-f_{2n}^2 \geq 0$$
By Cassini's Identity, the LHS is simply $(-1)^{2n} = 1$ which is always greater than zero, proving the original identity. The proof for even index terms decreasing follows the same route.