This seems too strong, even if $\omega$ is the empty set. Then $\chi_\omega = 0$ is in $BV(\Omega)$, and has total variation zero.
It therefore suffices to find a sequence $(f_n \mid n \in \mathbf{N})$ in $W^{1,1}(\Omega)$ so that $\lvert f_n \rvert_{L^1} \to 0$ but with total variation bounded below.
We construct Lipschitz functions $f_n$ on $\Omega = (-1,1)^2$ that are independent of $y$. Specifically let \begin{equation} f_n(x,y) =\begin{cases} x & \text{ on $[0,\frac{1}{2n}]$} \\ 1/n - x & \text{ on $[\frac{1}{2n},\frac{1}{n}]$} \end{cases} \end{equation} and extend this to be $\frac{1}{n}$-periodic in $x$.
Then $\lvert f_n \rvert_{L^\infty} \leq \frac{1}{2n} \to \infty$ but the total variation is constant equal to $\int_\Omega \lvert D f_n \rvert = 4$.