Let $f:\left[ a,b\right] \rightarrow \left[ a,b\right] $ be a continuous function. I want to prove that for all $\varepsilon >0$, there is $\delta >0,$ with the property:
(P) $\quad$ for all $x\in \left[ a,b\right] $ with $\left\vert x-f\left( x\right) \right\vert <\delta ,$ there is $u\in \left[ a,b\right] ,$ such that $$ f\left( u\right) =u \quad \text{ and }\quad \left\vert u-x\right\vert <\varepsilon . $$
By applying Brouwer's fixed point theorem, we deduce that $f$ has a fixed point $u_{0}\in \left[ a,b\right] .$
If $f$ is continuous, then $g\left( x\right) :=f\left( x\right) -x$ is uniformly continuous on $\left[ a,b\right] $ and it follows that $\forall \delta >0,$ $\exists \mu >0,$ such that $\forall x\in \left[ a,b\right] $ with $\left\vert x-u_{0}\right\vert <\mu ,$ $\left\vert g\left( x\right) -g\left( u_{0}\right) \right\vert <\delta ,$ i.e., $\left\vert x-f\left( x\right) \right\vert <\delta .$
How can I continue from now on? Any idea would be appreciated.
Let $$F= \{ u \in [a;b] | f(u)=u \}$$ denote the set of fixed points of $f$. We know that $F$ is a non empty closed set. For all $\varepsilon >0$ consider the set $$F_{\varepsilon} = \bigcup_{u \in F} B(u, \varepsilon)$$ This is the set of points whose distance from a fixed points is less than $\varepsilon$: this is an open set. Hence its complement is closed (and compact).
Then simply define $$\delta = \min_{x \notin F_{\varepsilon}} |f(x)-x|$$ and you are done. Indeed $\delta >0$, since the minimum exists and cannot be zero.
WHY DOES THIS WORK: if $|f(x)-x| < \delta$, then clearly $x$ does not satisfy the condition $x \notin F_{\varepsilon}$. In other words $x \in F_{\varepsilon}$, which means that there is a fixed point whose distance from $x$ is less than $\varepsilon$.