Is the following statement true. Assume that $f:[0,1]\to [0,1]$ is a continuous function such that $$\sup\_t\lim\sup\_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}<\infty,$$ then $f$ is Lipchitz continuous.



Criteria for Lipschitz continuity 
Even more is true: Suppose that $E$ is a measurable subset of an interval, $f$ is a function 
on $E$ so that for every $x\in E$, $|D^+(f)(x)|\le M$ (the quantity $|D^+(f)|$ is defined below). Then
$$
mes(f(E))\le M \cdot mes(E),
$$ 
see e.g. the proof of Lemma 3.13 in <http://www.ams.org/bookstore/pspdf/gsm-105-prev.pdf> 
("A First Course in Sobolev Spaces" by G.Leoni). Here $mes$ is the (outer) Lebesgue measure and 
$$
|D^+(f)(x)|= \lim sup\_{y\to x} \frac{|f(x)-f(y)|}{|x-y|} 
$$
Note that Lemma 3.13 assumes that $f$ is differentiable, but the proof uses only the upper bound $|D^+(f)(x)|\le M$ for all $x$. In your case, $E$ is an interval $[s,t]$ in $[0,1]$ and 
$$
M= \sup\_{x} \lim sup\_{y\to x} \frac{|f(x)-f(y)|}{|x-y|}
$$
Then, you conclude that for every $s$ and $t$ in the interval $[0,1]$, 
$$
|f(s)-f(t)|\le mes (f([s,t])) \le M |s-t| 
$$
and, hence, your function $f$ is $M$-Lipschitz. 


