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! :)

This is based on David's comment above. 
Choose $u,v, \epsilon$ such that $ a < u < v < b$ and $ \epsilon > 0 $. 
Let $$ S = \{ x \in [u,v] : f(x) + \epsilon x \geq f(u) + \epsilon u \}.$$ $S$ is not empty as $ u \in S$ and $S$ is closed as $f$ is continuous. 
Let $\sup S = t$, $t$ is in $S$ as $S$ is closed. We will show $t=v$.
If $ t < v$, then define for $ \delta \in (0, v- t] $, $g(\delta) = \sup_{0 < h \leq \delta} \dfrac{f(t+h) - f(t)}{h}$. 
Since $g(\delta)$ decreases to $\overline{D}f(t) \geq 0$  as $ \delta \to 0^+$, we must have $ g(\delta) > -\epsilon $ for sufficiently small positive $\delta$, and this means, the set $g(\delta)$ is a supremum of, must contain an element greater than $-\epsilon$. Hence there is a $t_1 \in (t,v]$ with $\dfrac{f(t_1) - f(t)}{t_1 - t} > -\epsilon$, we have $f(t_1) + \epsilon t_1 > f(t) + \epsilon t \geq f(u) + \epsilon u$, i.e., $t_1 \in S$. This contradiction implies $v = \sup S$ and $f(v) \geq f(u ) + \epsilon (v - u)$. Letting $\epsilon \to 0$, we have $f(v) \geq f(u)$. Hence $f$ is non-decreasing on $(a,b)$. This can be extended to all of $[a,b]$ by keeping $v$ fixed and letting $u \to a$ and keeping $u$ fixed and letting $v \to b$. 
