Let $$\mathbb{R}^2_+ = \{(x,y) \in \mathbb{R}^2 : y > 0\}$$ which has a boundary $B = \{(x,0) : x \in \mathbb{R}\}$.
Does the Poincare inequality hold on $H^1_0(\mathbb{R}^2_+)$: $$\lVert u \rVert_{L^2(\mathbb{R}^2_+)} \leq C\lVert \nabla u \rVert_{L^2(\mathbb{R}^2_+)}?$$
It passes the first test since the only constant in this space is $0$.
The two ways I know of proof do not work here. One uses the compact embedding of $H^1$ into $L^2$ which (I think) we don't have, and the second needs a domain bounded in one direction.
Take any $u\in C_0^1(\mathbb{R}^2_+)$ and define $$u_n(x)=u\left(\frac{x}{n}\right),\ \forall\ x\in \mathbb{R}^2_+. $$
Note that $$\int_{\mathbb{R}^2_+}|u_n(x)|^2dx=n^2\int_{\mathbb{R}^2_+}|u(x)|^2dx,$$
while $$\int_{\mathbb{R}^2_+}|\nabla u_n(x)|^2dx=n\int_{\mathbb{R}^2_+}|\nabla u(x)|^2dx.$$
Therefore, $$\|u_n\|_2^2=n^2\|u\|_2^2,\ \|\nabla u_n\|_2^2=n\|\nabla u\|_2^2.$$
If the Poincare inequality is true then, there is a positive constant $c$ such that $$n^2\|u\|_2^2\le c n\|\nabla u\|_2^2,\ \forall\ n,$$
however, this is an absurd.
Remark: This construction is valid on any open cone.