Prove a fact about a mapping of complete poset to itself

Question

Let M be a poset such that any its totally ordered subset has an upper bound: $\forall U \subseteq M$, $U$ is totally ordered $\exists m \in M: \forall u \in U \ \ u \leq m$. And let $f: M \to M$ be a mapping such that $\forall x \in M \ \ x \leq f(x) $ Prove that $\exists x_0 \in M: f(x_0)=x_0$

Looks a little scary for me, not sure how to approach the problem.

Answer 1

HINT: Let $U$ be a totally ordered subset of $M$ that is maximal with respect to inclusion, and let $x_0$ be an upper bound for $U$. (You will have to justify the existence of $U$.)

Answer 2

Hints:

What can be proved here on base of the lemma of Zorn?

If $m\in M$ is a maximal element then what can be concluded from $m\leq f(m)$?