Let $b$ be a continuous function defined on $[a,b]$. Suppose that there are $u,v \in \mathbb{R}$ such that the Dini number $D^{+}f$ satisfies $u \leq D^{+}f(x) \leq v$ for any $x \in [a,b]$. Show that $$uh \leq f(x+h)-f(x) \leq vh$$ for any $a \leq x < x+h \leq b$.
By definition
$$D^{+}f(x) = \limsup_{h \to 0+}\frac{f(x+h)-f(x)}{h}.$$
So,
$$u \leq \limsup_{h \to 0+}\frac{f(x+h)-f(x)}{h} \leq v.$$
I cannot develop any more. I'm probably not realizing some simple detail, since I'm not seeing where to apply the continuity of $f$. Could someone give me a hint?
As far as I can see, the answer is involved so I will give somewhat more than a hint but leave some work for you to do.
If $f$ were differentiable (all Dini derivatives equal) then this would follow easily from the mean value theorem (MVT) since there would exist $c \in (x , x+h)$ such that
$$\frac{f(x+h) - f(x)}{h} = f'(c)$$
and $u \leqslant f'(c) \leqslant v$ for all $c \in (a,b)$.
Working only with $D^+f$ we use a generalization of MVT by taking for $x \leqslant y \leqslant x+h$,
$$\phi(y) = f(y) - f(x) - \frac{f(x+h) - f(x)}{h}(y-x)$$
The continuity of $f$ implies the continuity of $\phi$ and, in addition, we have $\phi(x) = \phi(x+h)=0$. Ignoring the trivial case where $\phi$ is identically equal to $0$, there must be a point $c \in (x,x+h)$ where $\phi$ has a maximum or minimum. If it is a maximum or minimum, then we can use a well-known result that $D^+\phi(c) \leqslant 0$ or $D^+\phi(c) \geqslant 0$, respectively. This is proved in the solution to Exercise 21.6 in A Primer of Real Functions by Boas.
We also need the fact that
$$D^+\phi(y) = D^+f(y) - \frac{f(x+h)-f(x)}{h}$$
It is not generally true that $D^+(f +g) = D^+f + D^+g$, but it is true if $g'$ exists -- which is the case here.
Thus, if $\phi$ has a maximum at $c$ we have
$$D^+\phi(c) = D^+f(c) - \frac{f(x+h) - f(x)}{h} \leqslant 0,$$
which implies that
$$\frac{f(x+h)-f(x)}{h} \geqslant D^+f(c) \geqslant u,$$
giving us one side of the desired inequality. If there is also a point $c' \in(x,x+h)$ where $\phi$ has a minimum then we would have the second side as well,
$$\frac{f(x+h)-f(x)}{h} \leqslant D^+f(c') \leqslant v$$
Since we can't be sure that there are both maxima and minima in the open interval, another argument is needed to simultaneously obtain both sides of the inequality chain. For example if there is only a maximum point we have $\phi(y) > 0$ for $x < y < x+h$ and it follows that
$$\frac{f(x+h)-f(x)}{h} < \frac{f(y) - f(x)}{y-x} $$
I will leave the remainder of the argument to you.