This seems too strong, even if $\omega$ is the emptyregardless of which set. Then $\chi_\omega = 0$ is in$\omega \subset \Omega$ one works with. We suppose that $BV(\Omega)$, and$\omega$ has total variation zerobounded perimeter, so that $\chi_\omega \in BV(\Omega)$.
It therefore suffices to find a sequenceLet $(f_n \mid n \in \mathbf{N})$ be a sequence of functions in $W^{1,1}(\Omega)$ sowith \begin{equation} \lvert f_n \rvert_{L^1} \to 0 \text{ and } V(f_n) \geq 3 \, \mathrm{Per}(\omega). \end{equation} Let moreover $g_n \in W^{1,1}(\Omega)$ be a sequence of functions so that \begin{equation} \lvert g_n - \chi_\omega \rvert_{BV} \to 0; \end{equation} for example $\lvert f_n \rvert_{L^1} \to 0$ but with total variation bounded below$g_n$ can be obtained by a mollification argument. Then the two sequences combined have \begin{equation} \lvert f_n + g_n - \chi_\omega \rvert_{L^1} \to 0 \text{ but } V(f_n + g_n) \geq V(f_n) - V(g_n) \geq 3/2 \mathrm{Per}(\omega). \end{equation}
We construct Lipschitz functions $f_n$ on $\Omega = (-1,1)^2$$\Omega$ that are independent of $y$. Specifically let, given $A > 0$, \begin{equation} f_n(x,y) =\begin{cases} x & \text{ on $[0,\frac{1}{2n}]$} \\ 1/n - x & \text{ on $[\frac{1}{2n},\frac{1}{n}]$} \end{cases} \end{equation}\begin{equation} f_n(x,y) =\begin{cases} Ax & \text{ on $[0,\frac{1}{2n}]$} \\ A/n - Ax & \text{ on $[\frac{1}{2n},\frac{1}{n}]$} \end{cases} \end{equation} and extend this to be $\frac{1}{n}$-periodic in $x$.
Then $\lvert f_n \rvert_{L^\infty} \leq \frac{1}{2n} \to 0$$\lvert f_n \rvert_{L^\infty} \leq \frac{A}{2n} \to 0$ but the total variation is constant equal to $\int_\Omega \lvert D f_n \rvert = 4$$\int_\Omega \lvert D f_n \rvert = A \lvert \Omega \rvert$. To conclude it only remains to take $A > 3 \mathrm{Per}(\omega)/\lvert \Omega \rvert$.
