Proving a consequence of the Intermediate Value Theorem The following statement is said to be an immediate consequence of the Intermediate Value Theorem:
Let $I,J \subset \mathbb{R}$ be closed intervals satisfying $I \subset J$ and $J \subset f(I)$. Then $f$ possesses a fixed point in $I$. 
My approach: 
Let $ J = [a,b]$. Since $J \subset f(I)$ there exist $c,d \in I$ such that $f(c) = a$ and $f(d) = b$. 
At this point I do not really know how to continue although I think that I have proven a special case, which is the following:
Suppose that $I = [c,d]$. Define $h(x) = f(x) - x$. Since $I \subset J$ we have $h(c) = f(c) - c = a - c < 0$ and $h(d) = f(d) - d = b - d > 0$. Now the Intermediate Value Theorem can be applied and states that there is $\xi \in [c,d] = I$ such that $h(\xi) = 0$. That is $f(\xi) = \xi$ and therefore we would be finished. 
In general it is not clear that $I = [c,d]$, is it? 
Can anyone give a hint or help out proving the above statement? 

Your proof is alsmost complete. All you really need to check is what happens if $a=c$ or $b=d$ You didn't use the fact that $I=[c,d]$ anywhere in your proof. All you really used was the fact that $[c,d]\subseteq I$, which is true.
In particular:


*

*$h(c)=f(c)-c = a-c$ is true from the definitions of $h$ and $c$.

*$a-c<0$... well, almost. You actually have $a-c\leq 0$, but that's ok. I mean, it it's equal to $0$, then you're fine already, since $f(c)-c=0$ and $c$ is a fixed point.

*$h(d)=f(d)-d=b-d$ again, true from the definitions.

*$b-d>0$... same as above.

*Then, we apply the intermediate value theorem.

*Then, $h(\xi)=0$, and $\xi\in[c,d]$.


There is no point at which you used the fact that $I=[c,d]$.
