TITLE: What am I doing wrong in my proof?
QUESTION [5 upvotes]: If $f$ is continuously differentiable on $[a,b]$, then show that for all $x,y\in(a,b)$ using the definition of derivative
  $$
|f(x)-f(y)|\leq|x-y|\sup_{x\in[a,b]}|f'(x)|
$$

My attempt: Since $f'$ is continuous on $[a,b]$, it is bounded so there is $M>0$ such that $|f'(y)|\leq M$ for all $y\in[a,b]$. On the other hand, the extreme value theorem implies that $\sup_{y\in[a,b]}|f'(y)|$ exists so we can have
$|f'(y)|\leq\sup_{y\in[a,b]}|f'(y)|$ for all $y\in[a,b]$. Since $f$ is differentiable at $y$ by the definition of derivative I have
$$
f'(y)=\lim_{x\to y}\frac{f(x)-f(y)}{x-y}
$$
By using the precise definition of limit and some manipulations I get: for $\epsilon=1$ there exists $\delta>0$ such that 
$$
0<|x-y|<\delta\Longrightarrow|f(x)-f(y)|\leq(1+|f'(y)|)|x-y|
$$
But because of $|f'(y)|\leq\sup_{y\in[a,b]}|f'(y)|$ for all $y\in[a,b]$, I get
$$
0<|x-y|<\delta\Longrightarrow|f(x)-f(y)|\leq\left(1+\sup_{y\in[a,b]}|f'(y)|\right)
|x-y|
$$
which is different from what I was supposed to prove!

REPLY [3 votes]: If you keep dividing a closed interval repeatedly into two halves and pick one of the sub-intervals at each step the there will be  a unique point in the intersection. (The left end points form an increasing sequence and the right end points form an decreasing sequence, so they both have a limit and the lengths of the intervals tends to 0. Hence there is a unique point in the intersection).  If you are willing to use this fact then a proof can be given without any other theorem: suppose there exist $x,y$ such that $|f(x)-f(y)|>M|x-y|$ where $M>\sup \{|f'(x)|:a\leq x \leq b\}$. We may suppose $x<y$ Let $u= \frac {x+y} 2$. Then either $|f(x)-f(u)|>M|x-u|$ or $|f(u)-f(y)|>M|u-y|$. In the first case consider the interval $[x,u]$ and in the second case consider $[u,y]$. Repeating this process we get a decreasing sequence of intervals $[u_n,v_n]$ such that $|f(u_n)-f(v_n)|>M|u_n-v_n|$ for all $n$. Let $z$ be the unique point in the intersection of these intervals. Then $x_n \to z$ and $y_n \to z$. Since $\frac {f(u_n)-f(v_n)} {u_n-v_n} \to f'(z)$ we get a  contradiction by letting $n \to \infty$ in the inequality $|f(u_n)-f(v_n)|>M|u_n-v_n|$