To show that a recursively defined sequence by $x_1=\frac12$ and $x_{n+1}=\frac{x_n^3 + 2}{7}$ is Cauchy - How?

First, show by induction that, with $x_1=1/2$, it follows that $0<x_n\leq 1/2$. Then, since $(x^3 - y^3)=(x^2+xy+y^2)(x-y)$, we find that $$|x_{n+1}-x_{n}| = \frac{|x_n^3 - x_{n-1}^3|}{7}\leq\frac{\frac{1}{2^2} + \frac{1}{2^2}+\frac{1}{2^2}}{7} |x_{n}-x_{n-1}|=\frac{3}{28}|x_{n}-x_{n-1}|.$$ Now use the triangle inequality to estimate $|x_n-x_m|$. Can you take it from here?

You can prove by induction that $0<x_n< 1$ for all $n \in \mathbb N$.

The derivative of the function $f(x)=\frac{x^3+2}{7}$ is $f^\prime(x) = \frac{3}{7}x^2$. Hence $0\le f^\prime(x)\le \frac{1}{2}$ for $0 \le x \le 1$.

Using the mean value theorem, you get

$$\vert f(x)-f(y) \vert \le \frac{1}{2}\vert x -y \vert$$ for $x,y \in [0,1]$.

Which leads to

$$\vert x_{n+1}-x_n \vert \le \frac{1}{2} \vert x_n - x_{n-1} \vert $$ for all $ n \in \mathbb N$.

You then conclude that the sequence is Cauchy by induction.

This is a case of a more general result when the function is a contraction mapping.

By induction we see that $0<x_n<1$ for all $n$. Hence $|x_{m+1}-x_m| \leq \frac 3 7 |x_m-x_{m-1}|$ (by MVT theorem for $x \to x^{3}$). This gives $|x_{m+1}-x_m| \leq (\frac 3 7)^{m-1} |x_2-x_1|$. This implies that $\sum |x_{m+1}-x_m| <\infty$. Can you take it from here?