Let $f\in C^{BV}([0,1])$ (i.e. continuous and has bounded variation). Let the intervals $I$ and $T$ satisfy the following: $I\subset T\subset [0,1]$ and for sufficient small $\delta>0$, $\frac{|I|}{|T|}\leq \delta.$
Question: Is there an $\epsilon=\epsilon(\delta)>0$ such that
$$\frac{\underset{I}{Var}f}{\underset{T}{Var}f}\leq \epsilon.$$
As posed, the question has a trivial answer (as hinted by John in a comment): let $\epsilon=1$, then the inequality holds by the monotonicity of $\operatorname{Var}$.
A better question would be: is it true that for every $\epsilon>0$ there is $\delta=\delta(\epsilon)>0$ such that $$|I|\le \delta |T| \implies \operatorname*{Var}_I f\le \epsilon \operatorname*{Var}_T f \tag{?}$$
The answer is negative; the Cantor function is a counterexample. This function is of the form $f(x)=\mu([0,x])$ where $\mu$ is a singular measure. Since $ \operatorname*{Var}_I f = \mu(I)$ , the validity of (?) would imply that $\mu$ is absolutely continuous with respect to the Lebesgue measure, which is not the case.