Prove that a continuous function defined on an interval $[a,b]$ has a fixed point. I have to prove that :
Suppose that $f:[a,b] \to [a,b]$ is continuous. Prove that there is at least one fixed point in $[a,b]$.
But I don't know how to attack it since I can't apply anything of uniform continuity or other stuff, I have seen that it is recommended to use Bolzano's theorem but we haven't seen it yet. Thank you for your help.   
 A: If $f(a)=a$ we are finished. So we can assume that $f(a)\gt a$. Similarly, we can assume that $f(b)\lt b$.
Let $g(x)=x-f(x)$. Then $g(a)\lt 0$, and $g(b)\gt 0$. Also, $g$ is continuous. So by the Intermediate Value Theorem, we have $g(x)=0$ for some $x$ between $a$ and $b$. 
Remark: In a comment, OP has indicated that the Intermediate Value Theorem has not yet been proved in the course. 
One way around the problem is to prove the IVT! For guidance, maybe use this. 
A similar proof shows the existence of a fixed point directly. Start by observing that we can assume that $f(a)\gt a$ and $f(b)\lt b$.  Let $S$ be the set of all $x$ in our interval such that $f(x)\gt x$. Then $S$ is non-empty, and bounded above. Thus by the completeness of the reals, $S$ has a supremum $c$. Now show that $f(c)=c$. This can be done by showing that each of $f(c)\gt c$ and $f(c)\lt c$ leads to a contradiction. This is where one uses the continuity of $f$. 
A: Set $g(x) = f(x) - x$, thus $g(a) \geq 0$ and $g(b) \leq 0$. Also $g$ is 
continious. Now whether $g$ is constant zero function which proof completes or it 
has a negative or positive value in his range. Compare $g(c) < 0$ with $g(a)$ or $g(c) > 0$ with $g(b)$ and conclude (with intermediate value property) that in 
either cases $g$ vanish some where so at that point we must have $f(x) = x$.
If you don't want to use intermediate value theorem you must use this fact that 
$g$ (or any continous function) maps connected sets to connected sets.
A: Define the continuous function $g$ by
$$g(x)=f(x)-x$$
and convince yourself that
$$g(a)g(b)\le0$$
and then use intermediate value theorem.
A: let $f(x)$ be the continuos function , defined from$[a,b]→[a,b]$ and define another function $g(x)$=$x$ , now if $f(a)$ is greater than $a$ and $f(x)$ does not intersect $g(x)$ we can say $f(x)$>$x$ as if it is lesser and it does not intersect with $g(x)$ there will be drop , which creates discontinuity , thus $f(x)$>$x$ for $x$ belonging to [a,b] , but then $f(b)>b$ which creates contradiction , use similar argument for $f(a)<a$
