recursive sequence with continuous function 
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function, and let $\{x_{n}\}$ be a sequence defined by $x_{0}=0$ and $$x_{n+1}=f(x_{n})\quad(n\ge0).$$ Prove that if $\{x_{n}\}$ is bounded, then there exists $c\in\mathbb{R}$ such that $f(c)=c$.

Is it possible to show using the Bolzano-Weierstrass theorem?
Give some advice or hint. Thank you!
 A: If you suppose there is no fixed point for $f$, i.e. $f(x) \neq x$, for all $x\in \Bbb R$, then
$$x_1 = f(0) \neq 0.$$

If $x_1>0$ then $f(x) > x$ for all $x\in \Bbb R$ (because otherwise IVT would guarantee existence of a fixed point, and thus a contradiction). 
In general we have, therefore,
$$x_{n+1} = f(x_n) > x_n.$$
So $\{x_n\}$ is monotonically increasing and thus must have a limit, either finite or infinite. Suppose the limit is finite and equal to $\ell$. Then by Cauchy criterion, for large enough $n$, we would have
$$x_n-x_{n+1}<\varepsilon$$
that is
$$x_n -f(x_n)< \varepsilon.$$
This, however, implies that $\{x_n\}$ and $\{f(x_n)\}$ have the same limit, yielding
$$\{f(x_n)\} \to \ell,$$
Continuity of $f$ finally gives
$$f(\ell) = \ell,$$
a contradiction. 
So $\{x_n\}$ must diverge, and thus it is unbounded.

Similarly $x_1 <0$ gives a monotonically decreasing sequence $\{x_n\}$ that cannot converge to a finite limit, for similar reasons. 
A: Suppose $f$ has no fixed-point. Therefore we encounter one of two cases:
[Since $f$ is continous and we would otherwise get a fixed-point by the intermediate value theorem]
case 1: $f(x) < x \ \ \forall x \in \mathbb{R}$
In this case, $f$ has no lower bound (since $g(x) = x$ has none). Because $f(x) < x$ the sequence $(x_n)$ is strictly decreasing: $x_{k+1} = f(x_k) < x_k$ and $f(x_0) = f(0) < 0$.
Since the sequence is striclty decreasing it must have a limit $x \in \mathbb{R} \cup \{-\infty\}$.
If $x > -\infty$ we get
$$f(x) = f(\lim\limits_{n\to\infty} x_n) = \lim\limits_{n\to\infty} f(x_n) = x$$ since f is continous.
This contradicts our assumption that $f$ has no fixed point.
Hence $x_n \to -\infty$ and therefore $\{x_n\}$ is unbounded.
case 2: $f(x) > x \ \ \forall x \in \mathbb{R}$
Similarly $f$ now has no upper bound. $(x_n)$ is now strictly increasing.
In both cases we conclude:
$$f\ \text{has no fixed point} \Rightarrow \{x_n\}\ \text{is unbounded}$$
A: Although I find the already given answers more interesting, I will add this proof for sake of completeness.
Yes, it is possible to use Bolzano-Weierstarss theorem.
If $(x_n)$ is a real valued and bounded sequence, then it has a convergent subsequence $(x_{n_j})$ and we denote $\lim\limits_{j\to\infty}x_{n_j}:=c$.
It follows from continuity that $c=\lim\limits_{j\to\infty}x_{n_j}=\lim\limits_{j\to\infty}f(x_{n_{j-1}})=f(\lim\limits_{j\to\infty}x_{n_{j-1}})=f(c)$. Hence, the sequence has a fixpoint.
