Not an optimal one necessarily, but there does exist a non-trivial constant for any bounded subset $\Omega\subset \mathbb{R}^n$. 

Let $u\in W_0^{1,p}(\Omega)$, with $\Omega$ bounded. The classical Poincare inequality is

$||u||_{L^p(\Omega)}\leq C||\nabla u||_{L^p(\Omega)}$. In the case of $p=2$ , the optimal constant is given by $\frac{1}{\lambda}$ where $\lambda$ is the first eigenvalue to $\begin{cases} \Delta u=\lambda u \hspace{5mm}\forall x\in \Omega \\ u=0 \hspace{12mm}\forall x\in bd(\Omega) \end{cases}$

For convex domains, for $p=1$, we have that $C=\frac{diam(\Omega)}{2}$, and for $p\geq 2$  we have $C=(\frac{\pi_p}{d})^p$, where $\pi_p=(\frac{2\pi (p-1)}{psin({\pi/p})})^{1/p}$

There has been recent work done on finding a non-trivial Poincare constant for $u\in W^{1,p}$ (where $u$ doesn't necessarily vanish at the boundary) for not only non-convex subsets $\Omega$, but more generally any open & bounded subset (not necessarily connected). The paper is here: https://arxiv.org/pdf/1811.07470.pdf

The idea of the paper is straightforward: every open and bounded $\Omega$ can be covered by dyadic cubes. Non-trivial local estimates on the cubes are derived, and then extended to the entire $\Omega$. The result(s) extend to a totally disconnected $\Omega$, as there are non-trivial examples of totally disconnected sets with positive measure.

Another interesting observation is how the author proves a lemma which gives necessary and sufficient conditions for $u\in L^p_{loc}(\Omega)$ to also be $u\in L^p(\Omega)$.