Proof: If $f'=0$ then is $f$ is constant I'm trying to prove that if $f'=0$ then is $f$ is constant WITHOUT using the Mean Value Theorem.
My attempt [sketch of proof]: Assume that $f$ is not constant. Identify interval $I_1$ such that $f$ is not constant. Identify $I_2$ within $I_1$ such that $f$ is not constant. Repeat this and by the Nested Intervals Principle, there is a point $c$ within $I_n$ for any $n$ such that $f(c)$ is not constant... This is where I realized that my approach might be wrong. Even if it isn't I don't know how to proceed.
Thanks for reading and any help/suggestions/corrections would be appreciated.

So we have to prove that $f'(x)\equiv0$ $\ (a\leq x\leq b)$ implies $f(b)=f(a)$, without using the MVT or the fundamental theorem of calculus.
Assume that an $\epsilon>0$ is given once and for all. As $f'(x)\equiv0$, for each fixed $x\in I:=[a,b]$ there is a neighborhood $U_\delta(x)$ such that
$$\Biggl|{f(y)-f(x)\over y-x}\Biggr|\leq\epsilon\qquad\bigl(y\in\dot U_\delta(x)\bigr)$$
($\delta$ depends on $x$). For each $x\in I\ $ put $U'(x):=U_{\delta/3}(x)$. Then the collection $\bigl(U'(x)\bigr)_{x\in I}$ is an open covering of $I$. Since $I$ is compact there exists a finite subcovering, and we may assume there is a finite sequence $(x_n)_{0\leq n\leq N}$ with
$$a=x_0<x_1<\ldots< x_{N-1}<x_N=b$$
such that $I\subset\bigcup_{n=0}^N\  U'(x_n)$. The $\delta/3$-trick guarantees that $$|f(x_n)-f(x_{n-1})|\leq \epsilon(x_n-x_{n-1}).$$ By summing up we therefore obtain the estimate $|f(b)-f(a)|\leq \epsilon(b-a)$, and as $\epsilon>0$ was arbitrary it follows that $f(b)=f(a)$.
