Continuous function on $[a, b]$ with bounded upper and lower derivatives on $(a, b)$ is Lipschitz. This question was posted previously, but without the critical assumption that $f$ be continuous, and so this question was left unanswered. 
I am using the following definitions:
Upper derivative of $f$ at $x$:
$\overset{-}{D} \, f(x) = \displaystyle \lim_{h \to 0} \left [\displaystyle\sup_{0 < |t|\leq h} \frac{f(x + t) - f(x)}{t} \right ]$
Lower derivative of $f$ at $x$:
$\underset{-}{D} \, f(x) = \displaystyle \lim_{h \to 0} \left [\displaystyle\inf_{0 < |t|\leq h} \frac{f(x + t) - f(x)}{t} \right ]$.
My idea is that if, say, $\overset{-}{D}$ is bounded then the $\sup$ in the definition is finite call it $C$, and we can use this as the constant in the Lipschitz condition. Similarly for $\underset{-}{D}$; call its bound $C'$. However, if $\underset{-}{D} \neq \overset{-}{D}$ then $f$ is not differentiable, then these bounds are different. In this case, can I just take $\textrm{max} (|C|, |C'|)$ as the Lipschitz constant?
Or am I way off?
PS: as a side question, why do we need $t$ in absolute value in the definitions? Is this because we want to approach from the left as well?
 A: This solution uses "$\underline{D}f \geq 0 \Rightarrow f$ increasing". (This is another exercise in Royden, no. 19, preceding the one you're asking about. Fairly easy to show, and there's another stackexchange thread detailing the solution I believe.) 
Let $\overline D \leq \alpha$, $\underline D f \geq \beta$, be the finite bounds on the upper and lower derivative. The idea is simple. Define increasing and decreasing functions, and find bounds based on those functions.
$$ g:= f + \beta x $$
$$ h:= f - \alpha x $$ 
Observe that adding the linear terms raises the lower bound on the lower derivative of $g$, and respectively reduces the upper bound of the upper derivative of $h$. 
$$ \underline D g  \geq 0,$$
so $g$ increasing, so $f(u) - f(v) \leq \beta (u-v)$, if $v < u$, for arbitrary $v,u \in [a,b]$. Similarly, you can verify that $f(v) - f(u) \leq \beta (u-v)$, if $u<v$.
We can see that $\overline Dh \leq 0$, and we know that $\underline D (-h) = -\overline D h$ (simple to check), so $\underline D(-h)\geq 0$ and $-h$ is increasing.
From this, we can get the following inequalities:
$$ f(v) - f(u) \leq \alpha (v-u), \text{ if } v>u.$$
$$ f(u) - f(v) \leq \alpha (v-u), \text{ if } v<u.$$
W.l.o.g., let $\alpha \geq 0$, since otherwise you can apply this method to $-f$, and $-f$ has an upper bound on the upper derivative that is positive, because it is $\beta \leq  \alpha < 0$.
Now combining the four inequalities and the facts that $\alpha \geq 0$, $\alpha \geq \beta$, for each case $u<v$ or $u>v$ separately, gives us that $|f(v)-f(u)|\leq \alpha |v-u|$. $u,v$ were arbitrarily chosen, so we are done.
A: Since lower and upper derivatives are bounded, then there exists $M > 0$ such that
\begin{align*}
  -M \leq \underline{D}f(x)\leq \overline{D}f(x)\leq M,\quad \forall x\in [a,b].
 \end{align*}
Recall
\begin{align*}
 \overline{D}f(x) = \lim_{h\rightarrow 0^+}\left(\sup_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}\right)
 \quad\text{and,}\quad
 \underline{D}f(x) = \lim_{h\rightarrow 0^+}\left(\inf_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}\right)
 \end{align*}
Claim 1 For every $x\in (a,b)$, there exists an open interval $I\subseteq[a,b]$ such that $x\in I$ and $|f(y)-f(x)|\leq M|y-x|\}$ for all $y\in I$.
Proof of Claim 1. Let $x\in [a,b]$ and $\varepsilon > 0$. Then there exist $h_0$ and $h_1$ such that $h\leq \min\{h_0,h_1\}$ implies
\begin{align*}
  \sup_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}< M  +\varepsilon,
  \quad\quad
  \inf_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}> -M-\varepsilon.
 \end{align*}
It further implies that
\begin{align*}
  \left|{\frac{f(x+t)-f(x)}{t}}\right| <M + \varepsilon,\quad\forall |{t}| \leq h.
 \end{align*}
We can also write it as $|f(y)-f(x)| \leq M|{y-x}|$ for all $|{y-x}| \leq h$ and $y\in [a,b]$.
Let $I = (x-h, x+h)\cap (a,b)$, then $I$ is the desired open interval.
Claim 2 For every $[c,d]\subseteq[a,b]$, we have $|{f(c)-f(d)}|\leq M |{c-d}|\}$.
Proof of Claim 2. Define
\begin{align*}
  S = \{x\in[c,d]:|{f(c)-f(x)}|\leq M |{c-x}|\}.
 \end{align*}
Since $c\in S$, then $S$ is nonempty. Let $u = \sup S$. Since $f$ is a continuous function, then $S$ is closed. Hence, $u\in S$ and $u\leq d$. Suppose $u\neq d$, that is, $ u < d$. From Claim 1, there exists an open interval $I$ containing $u$. Let $y\in I$ with $u < y < d$. Then
\begin{align*}
  |{f(y) - f(c)}| \leq |f(y)-f(u)| + |f(u)-f(c)|\leq M(y-u) + M(u-c) = M(y-c).
 \end{align*}
Therefore, $y\in S$ and $y > u$, which contradicts $u = \sup S$. Thus, $u = d$.
Suppose there exists $x,y\in [a,b]$ such that $|{f(x)-f(y)}| > M|{x-y}|$, which contradicts Claim 2, since $[x,y]\subseteq [a,b]$.
