If $S$ is is a sequence of intervals over the reals starting with $S_0 = (a,b)$ where $a,b\in\mathbb{R}$
and the rest are generated by some function $f : S_n → S_{n+1}$
If I can show that:
$S_{n+1} \subsetneq S_n \forall\, n ≥ 0$
Does this also mean that if $x_{n+1} = f(x_n)$ then for some $x_0\in S_0$ the $\lim\limits_{n\to \infty} x_n$ exists?
No. Take the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by
$$\forall x \in \mathbb{R}, \quad f(x)=\left\lbrace\begin{array}{ll} -\frac{1}{2}x+4 & \text{if } x \leq 2 \\ -x+5 & \text{if } 2 < x < 3 \\ -\frac{1}{2}x + \frac{7}{2} & \text{if } x \geq 3\end{array}\right.$$ Define $S_0= ]0,5[$, and let $S_{n+1} = f(S_n)$ for every $n \geq 0$.
It is quite clear that $S_{n+1} \subsetneq S_n$ for every $n \geq 0$.
But $f(2)=3$ and $f(3)=2$ so the sequence $x_{n+1}=f(x_n)$ with $x_0=2$ does not converge.