Prove that a recurrent sequence $u_n$ is convergent if $\lim \limits_{n \to \infty}(u_{n+1}-u_n)=0$ Let $f:[a;b] \to [a;b]$ be a continuous function ($a$ and $b$ real numbers) and $(u_n)$ a sequence defined by  $u_0 \in [a;b]$ and $u_{n+1}=f(u_n)$
Prove that if $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$ then $(u_n)$ converges
I tried using subsequences and the bolzano weierstrass theorem but couldn't even get close.
Any help would be welcome.
 A: Lemma : If $(u_n)$ is a sequence such that $u_{n+1} - u_n \rightarrow 0$, then the set of all the adherence value (the word in french is "valeur d'adhérence") is an interval.
Proof : Let $x$ and $y$ two different adherence value (if they exist). Let $c\in ]x,y[$. To show it's an adherence value of $u$, we just need to show that $$\forall \varepsilon >0, \forall N\in \mathbb{N}, \exists n\geqslant N, |u_n -c| \leqslant \varepsilon$$
We take $\varepsilon >0$. Since $u_{n+1} - u_n \rightarrow 0$, there exists $N$ such that for all $n\geqslant N$, $|u_{n+1} - u_n|\leqslant \varepsilon$. Since $x$ is an adherence value of $u$, with $x<c$, there exists a rank $p\geqslant N$ such that $u_p <c$. Since $y$ is also one with $y>c$, there exists $q>p$ such that $u_q >c$. Let's consider the smallest $n>p$ such that $u_n >c$. Then we have $$u_n >c, u_{n-1}\leqslant c \text{ and } |u_n-u_{n-1}|\leqslant \varepsilon$$
So $u_n$ is in $]c, c+\varepsilon]$, which gives the result
Proof of the theorem :
We first note that all adherence values are fixed points of $f$.
From the Lemma, we get that if we have $2$ adherence values, for instance $x$ and $y$, then $[x,y]$ is only consituted of adherence values, and so of fixed points.
That means since we have to go from $x$ to $y$ with jumps that tend to $0$, when the length of the jumps becomes inferior strictly to $|y-x|/2$, then the sequence take a value in $[x,y]$, so takes a fixed point as value, so that the sequence gets constant after this rank.
That contradicts the fact that $x$ and $y$ are adherence values.
So $(u_n)$ converges.
A: $\{u_n\}$ is a sequence taking values in the compact segment $[a,b]$. Hence it has a convergent subsequence $\{u_{\varphi(n)}\}$ where $\varphi : \mathbb N \to \mathbb N$ is strictly increasing. Let say that $\lim\limits_{n \to \infty} u_{\varphi(n)} = l \in [a, b]$.
Suppose that $\{u_n\}$ has a second limit point $L \neq l$. We can find a subsequence $\{u_{\psi(n)}\}$ converging to $L$. Without loss of generality, we can suppose $l \lt L$. As $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$, any $ x \in [l , L]$ is a limit point of $\{u_n\}$.
Now as $f$ is supposed to be continuous, we get that $f(x) = x$ for all $x \in [l,L]$. But if $u_N \in (l,L)$ for $N \in \mathbb N$, we get the contradiction $u_p = u_N$ for all $p \ge N$ and therefore that $u_N$ is the only limit point of $\{u_n\}$.
Conclusion: $\{u_n\}$ has a single limit point and therefore converges.
