Right derivative of continuous function nonnegative implies increasing function?
See snapshot of Hardy's book question 19: 
As pointed out by Paramanand in the comments, without continuity no conclusion on monotonicity at right neighborhoods of points in the domain can be drawn, even though $$f'_+(x) = \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h} \geq 0$$ $\forall x\in\Bbb R$. I show in this answer, too, for completeness, a counterexample that I already posted here. It's a function $f(x)$ with a well defined, non negative right derivative in $\Bbb R$, which however has $f(0) = 0$ and $f(x) <0$ for $x>0$.
The red lines are parabolic envelopes with equations $y=-x|x|$ and $y=-\frac{1}{2}x|x|$ and the function is $$f(x) = \begin{cases}\frac{\sqrt{2^{-k}}-\sqrt{2^{-k+1}}}{\sqrt[4]{2^{-k+1}}-\sqrt[4]{2^{-k}}}(x+\sqrt[4]{2^{-k+1}})- \sqrt{2^{-k+1}} & \left(-\sqrt[4]{2^{-k+1}}\leq x<-\sqrt[4]{2^{-k}}; \ k\in \Bbb Z\right)\\ 0 & (x=0)\\ \frac{\sqrt{2^{-k}}-\sqrt{2^{-k+1}}}{\sqrt[4]{2^{-k+1}}-\sqrt[4]{2^{-k}}}(x-\sqrt[4]{2^{-k+1}})+ \sqrt{2^{-k+1}}& \left(\sqrt[4]{2^{-k}}\leq x<\sqrt[4]{2^{-k+1}}; \ k\in \Bbb Z\right) .\end{cases}$$
Let us now demonstrate the following
Theorem Let $f(x)$ be a continuous function in $[a,b]$ such that \begin{equation}f'_+(x) = \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h} \geq 0, \ \ \forall x\in[a,b).\tag{1}\label{eq:1}\end{equation} Then, for all $a\leq x \leq y\leq b$, $f(y)\geq f(x)$.
We will need the following variation of Rolle's Theorem.
Lemma If $f(x)$ a is continuous function in $[a,b]$ with well defined right derivative $f'_+(x)$, and such that $f(a) = f(b)$, then there are points $\alpha,\beta\in[a,b]$ for which $f'_+(\alpha)\geq 0$ and $f'_+(\beta)\leq 0$.
Since $f(x)$ is continuous in $[a,b]$, by Weierstrass Theorem it must reach its minimum and maximum in such interval. If $f(a) = f(b)$ is minimum and maximum then the function is constant and the Lemma is proved. If $f(a) = f(b)$ is a minimum, then the maximum must be $\beta \in (a,b)$, and we have $f(x) \leq f(\beta)$. Hence, if $x\in(\beta, b] $, then we obtain $\frac{f(x)-f(\beta)}{x-\beta}\leq 0$. Therefore $f'_+(\beta) \leq 0$. Also, by minimality of $a$, we have $f(x) \geq f(a)$ for $x\in [a,b]$ so that $f'_+(a) \geq 0$. Similary, we can prove the statement if $f(a) = f(b)$ is a maximum, and if both minimum and maximum are in the open interval $(a,b)$.$\square$
Now we are ready to prove the Theorem, by contradiction. Suppose there are $c_1, c_2\in[a,b]$ such that $c_1 < c_2$ and $f(c_1) > f(c_2)$.
As when proving the Mean Value Theorem, consider an auxiliary function $$F(x) = f(c_2)-f(x) + K(x-c_2),$$ with $$K = \frac{f(c_2)-f(c_1)}{c_2-c_1}< 0.$$ It is easy to verify that $F(x)$ satisfies the Lemma hypotheses in $[c_1,c_2]$, with $F(c_1) = F(c_2) = 0$, and \begin{equation}F'_+(x) = K-f'_+(x).\tag{2}\label{eq:2}\end{equation} By the Lemma there is a point $c\in[c_1,c_2]$ such that $$F'_+(c) \geq 0,$$ which, by \eqref{eq:2}, yields $$f'_+(c) \leq K < 0,$$ a contradiction. $\blacksquare$