Is this proof redundant? In the proof of :
"Let $f : I \to \mathbb{R}$, a continuous function, let $x_* \in ]a, b[$ a local extremum of $f$, then $f'(x_*) = 0$"  
There is a point where, for the case of a local maximum, we take $\epsilon > 0$ small enough so that:  


*

*$[x - \epsilon, x +\epsilon] \subset ]a,b[$

*$f(x) < f(x_*)$ for all $x \in [x_* - \epsilon, x_* + \epsilon]$


But aren't those two points redundant? Since $[x - \epsilon, x +\epsilon] \subset ]a,b[$ and x is already the local maximum of $]a,b[$ doesn't it mean that any value of ]a,b[ is necessarily inferior to $x_*$? 
Additional details about my reasoning, hard to make it clear but in case that helps:
As I am asking this question, I suppose that there is indeed some cases where a function would happen to be constant on an interval and where any value of this interval would be a local maximum; in this case — coming back to the demonstration — we would have values of epsilon where $f(x)$ wouldn't be strictly inferior to $f(x_*)$.
It would mean that the proof ins't redundant, but then wouldn't it be contradictory with $\epsilon$ being small enough? 
Thank you
 A: You have also to assume that $f$ is differentiable at $x_*$. For instance, the function $f(x)=-|x|$ has a local maximum at $0$, but has no derivative at that point.
Note that a function such as $f(x)=3x-x^3$ has a local maximum at $1$, but it is false that $f(x)<f(1)$ for every $x$.
A point $x_*$ is a local maximum for $f$ defined over some set $D\subseteq\mathbb{R}$ if there exists $\delta>0$ such that, for $x\in \mathopen]x_*-\delta,x_*+\delta\mathclose[\cap D$, $f(x)\le f(x_*)$.
In the case of $D=\mathopen]a,b\mathclose[$, you have
$$
\mathopen]x_*-\delta,x_*+\delta\mathclose[\cap\mathopen]a,b\mathclose[=
\mathopen]x_*-\delta',x_*+\delta'\mathclose[
$$
for some $\delta'>0$ and, taking $\varepsilon=\delta'/2$, we have
$$
[x_*-\varepsilon,x_*+\varepsilon]\subset\mathopen]a,b\mathclose[
$$
and $f(x)\le f(x_*)$, for every $x\in[x_*-\varepsilon,x_*+\varepsilon]$.
One needs to restrict the study to $[x_*-\varepsilon,x_*+\varepsilon]$ (I'd prefer an open subinterval, though) exactly because it is not generally true that a local maximum is an absolute maximum.
The usual definition of local maximum just requires $f(x)\le f(x_*)$ (not a strict inequality), but it's a matter of conventions. If your textbook enforces the strict inequality, then a function that's constant on a subinterval of the domain hasn't a local maximum at those points.
