Fix-Point Theorem Proof. Firstly, the assignment:

Let $a,b \in\mathbb{R}$ and $a < b$. Furthermore let $f: [a,b] \rightarrow [a,b]$ be monotone increasing. Show that if $x:= \mathbf {sup}\{y \in [a,b] \| \ y ≤ f(y)\}$ then $f(x) = x$.

It's good on me that the assignment itself actually tells you what $x$ is so I can prove the theorem by showing that for $\forall x' \not= x: f(x) \not= x$ and I'm done. However, I have not arrived at a contradiction so far.
Any tips?
 A: What's below is pretty much an adaptation of the Knaster-Tarski Fixpoint Theorem's proof.

You should prove that $x$ is well defined, that $x\in [a,b]$ and that $\{y \in [a,b] \colon \ y ≤ f(y)\}\neq \varnothing$.
I'll prove that $x\leq f(x)$ and $f(x)\leq x$.


*

*$\boxed{x\leq f(x):}$ To prove this it's sufficient to prove that $f(x)$ is an upper bound of $\{y \in [a,b] \colon \ y ≤ f(y)\}$, because, by definition $x$ is the smallest of these upper bounds. So take $y\in \{y \in [a,b] \colon \ y ≤ f(y)\}$, by definition of $x$ it holds that $y\leq x$, thus, since $f$ is increasing, it follows that $f(y)\leq f(x)$ and since $y\leq f(y)$, one gets $y\leq f(x)$. From the fact that $y$ was an arbitrary element of $\{y \in [a,b] \colon \ y ≤ f(y)\}$, it follows that $f(x)$ is an upper bound of this set and therefore $x\leq f(x)$.

*$\boxed{f(x)\leq x:}$ Since $x\leq f(x)\implies f(x)\leq f(f(x))$, it follows that $f(x)\leq f(f(x))$, which means that $f(x)\in \{y \in [a,b] \colon \ y ≤ f(y)\}$, so, by definition of $x$, it comes $f(x)\leq x$.


$\therefore f(x)=x$.
A: Hint: If $x=b$ you're done. Else try to consider two sequences $a_n\in[a,x)$ and $b_n\in(x,b]$ both converging to $x$.
A: The set $Y=\{y\in[a,b]\mid y\le f(y)\}$ you take the supremum of is nonempty (it contains $a$) and a subset of $\left[a,b\right]$. Therefore $x\in\left[a,b\right]$. 
If $f(x)<x$, there exists $y\in \left(f(x),x\right)\cap Y$; from $x>y$ we obtain $f(x)\ge f(y)\ge y>f(x)$, contradiction. 
Therefore $f(x)\ge x$, hence $f\left(f(x)\right)\ge f(x)$ so that $f(x)\in Y$ and hence $x\ge f(x)$. We conclude $f(x)=x$.
A: I think it is more suggestive if we set 
$ A=\{y \in [a,b] \| \ y ≤ f(y)\}$.
The fact that $x$ is the sup of $A$ and $A$ is defined by inequality hint that we should prove $ x\le f(x)$ and $x\ge f(x)$
i) $ x\le f(x)$ 
We approximate $x$ by a sequence in $A$. However, $f$ is not supposed to be continuous but monotone,we have to choose a monotone sequence!
ii) There is no inequality suggests the reverse direction, so we should use contradiction argument: suppose that $x<f(x)$. The only thing that we can think of is the monotonicitassumption; to use it, we  set $z=f(x)$. Then $z=f(x)\le f(z)$. Hence, $z\in A$ but this is impossible since $z>x=\text{sup}A$.
