Let $f:\mathbb{R}\to (0,+\infty)$ be a continuously differentiable function on $(-\epsilon, 1 + \epsilon)$ for some $\epsilon > 0$ and piecewise smooth on the rest of $\mathbb{R}$ which is non-increasing on the interval $[0,1]$. Let $a,b,c\in\mathbb{R}$ such that $0 < a < b$ and $0 < c$. Define $g(x) = \frac{-f'(x)}{f(x)}$ and
$$A = \{x\in [0,1]:g(x) > c\}$$
$$B = g^{-1}(c/2,+\infty)$$
$$C = \{x\in A: f(x)\in [a, b)\}$$
where $g^{-1}(c/2,+\infty)$ is taken as the pre-image of $g$ w.r.t. the interval $(c/2,+\infty)$. Is it then true that $C$ is covered by a single connected component of $B$? Our assumptions imply that $C\subset A\subset B$ and that $f$ is strictly decreasing and hence injective on $A$. But other than that I don't really know how to prove or diprove this claim.
No. For instance, consider an $f$ which has $f'(0)=f'(1)=-1$ and $f'(1/2)=0$. Then for small $c>0$, $A$ and $B$ will contain $0$ and $1$ but not $1/2$, and $a$ and $b$ can be chosen such that $C$ is all of $A$ and so is not covered by any single connected component of $B$. (If you want to require $f'<0$ on $[0,1]$, you can still achieve that by having $f'$ get very close to $0$ at $1/2$, so that $1/2$ will still not be in $B$ for an appropriate value of $c$.)