A function with a non-negative upper derivative must be increasing? I am trying to show that if $f$ is continuous on the interval $[a,b]$ and its upper derivative $\overline{D}f$ is such that $ \overline{D}f \geq 0$ on $(a,b)$, then $f$ is increasing on the entire interval. Here $\overline{D}f$ is defined by
$$
\overline{D}f(x) = \lim\limits_{h \to 0} \sup\limits_{h, 0 < |t| \leq h} \frac{f(x+t) - f(x)}{t} 
$$
I am not sure where to begin, though. Letting $x,y \in [a,b]$ be such that $x \leq y$, suppose for contradiction that $f(x) > f(y)$, then continuity of $f$ means that there is some neighbourhood of $y$ such that $f$ takes on values strictly less than $f(x)$ on this neighbourhood. Now I think I would like to use this neighbourhood to argue that the upper derivative at $y$ is then negative, but I cannot see how to complete this argument.
Any help is appreciated! :)

Probably not the best approach, but here is an idea: show taht MVT holds in this case:
Lemma Let $[c,d]$ be a subinterval of $[a,b]$. Then there exists a point $e \in [c,d]$ so that 
$$\frac{f(d)-f(c)}{d-c}=\overline{D}f(e)$$
Proof:
Let $g(x)=f(x)-\frac{f(d)-f(c)}{d-c}(x-c) \,.$
Then $g$ is continuous on $[c,d]$ and hence it attains an absolute max and an absolute minimum. Since $g(c)=g(d)$, then either $g$ is constant, or one of them is attatined at some point $e \in (c,d)$. 
In the first case you can prove that $\overline{D}g=0$ on $[c,d]$, otherwise it is easy to conclude that $\overline{D}g(e)=0$.
Your claim follows immediately from here.
