No differentiable function on $[-1,1]$ with derivative $\operatorname{sgn}(x)$ How can I prove for certain that there exists no differentiable function $g(x)\colon [-1,1]\rightarrow \mathbb{R}$ with
$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1,  & \text{for $x > 0$} \\
0, & \text{for $x=0$} \\ -1, & \text{for $x<0$} \end{cases} $$
I'm currently jumping to the absolute value function but that's not differentiable at $x = 0$ and that doesn't exactly disprove the existence of others.
Looking at the limits
\begin{align}
\lim_{h\to 0^{+}} g'(h) &=1\\
&= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\
&\neq  \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\
&= -1\\
&= \lim_{h\to 0^{-}}g'(h)\\
&\neq 0\\
&= g'(0)
\end{align}
which implies discontinuity at $h = 0$.
Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?

Suppose that such a function $g$ existed.  Then based on your definition of $g'$ we can conclude that
$$g(x) = \begin{cases}
-x+c_1, &x<0\\
a, &x = 0\\
x+c_2, &x > 0
\end{cases}$$ where we don't know the value of $a$.  The only question is whether or not there are some choice of constants to make $g'(x) = \operatorname{sgn}(x)$.  If that were to work, then following would work out:
$$0 = g'(0) = \lim_{h\to 0}\frac{g(h) - g(0)}{h}.
$$
Because the definition of $g$ differs between the left and right of $0$, we need to consider the left and right limits, though as we will see it suffices to consider the right-hand limit:
$$\lim_{h\to 0^{+}}\frac{g(h) - g(0)}{h} = \lim_{h\to 0}\frac{h+c_2 - c_2}{h} = \lim_{h\to 0}\frac{h}{h} = 1 \ne 0.$$  Thus, we can conclude that there is no such function $g$ with $g'(x) = \operatorname{sgn}(x)$.
