This is exercise 12.25 from Leoni's book.
Update: @Lukas Geyer has provided a counterexample for the case that $\Omega$ has a very bad boundary, which suggests that original exercise might be wrong. However, I am still wondering that if we assume $\Omega$ has smooth boundary, could the result be hold? i.e., we assume that $\Omega$ has smooth boundary but still not bounded, only finite.
Given $\Omega\subset R^N$ is an open finite set, i.e., $|\Omega|<\infty$. But $\Omega$ may not have a nice boundary or being bounded, i.e., the usual Sobolev Embedding won't work on this set. Now I am trying to prove that given any $u\in W^{1,p}(\Omega)$, where $1<p<\infty$, I have for any $1< q<p$, that $$\left(\int_\Omega|u-u_\Omega|^q\right)^{\frac{1}{q}}\leq C\left(\int_\Omega|\nabla u|^p\right)^\frac{1}{p} \,\,\,\,\,\,\,\,\,\,\,\,(1)$$ where $C$ depends on $p$, $q$, and $\Omega$.
To do so, I think I need to use an exercise which I proved before: for $\Omega$ open finite, the space $W^{1,p}(\Omega)$ is compact embedded in $L^q(\Omega)$ for any $q<p$.
Next, I proceed as usual, suppose $(1)$ does not hold and I have $u_n\in W^{1,p}$ such that
$$\left(\int_\Omega|u_n-(u_n)_\Omega|^q\right)^{\frac{1}{q}}\geq n\left(\int_\Omega|\nabla u_n|^p\right)^\frac{1}{p} \,\,\,\,\,\,\,\,\,\,\,\,(2)$$
Next, by defining $$v_n:=\frac{u_n-(u_n)_\Omega}{\|u_n-(u_n)_\Omega\|_{L^p(\Omega)}} $$ we have $\|v_n\|_{L^p}\equiv 1$ and $\|\nabla v_n\|_{L^p}\to 0$.
Then, if we have $W^{1,p}$ is compact embedded in $L^p$ we would be done already and this is how usual Poincare proved. However, here we only have $W^{1,p}$ is compact embedded in $L^q$, thus, we only have $v_n\to v_0$ strongly in $L^q$ for some $v_0$ such that $\nabla v_0=0$. If we can prove $v_0$ is not $0$, then we done. However, since generally we have $\|v_n\|_{L^q}\leq \|v_n\|_{L^p}$ if $q<p$, I can not rule out the probability that $v_0=0$...
Any help would be really welcome!
This is not true without some additional assumptions on the boundary. Just to make life simple, the following example uses $N=2$, but it is easy to modify for any $N \ge 2$.
Fix $p > q \ge 1$. Let $\Omega$ be a domain in the plane which consists of squares $(Q_k)_{k=0}^\infty$, of side length $2^{-k}$ centered at points on the $x$-axis, together with horizontal strips $(S_k)_{k=1}^\infty$, symmetric with respect to the $x$-axis, of width $2^{-k} t_k$ and height $2^{-k}t_k^{-1}$, joining $Q_{k-1}$ to $Q_k$, where $t_k \to \infty$ is determined below. By construction, the area of $\Omega$ is finite, since the areas of both $Q_k$ and $S_k$ are $4^{-k}$.
Now fix $n \ge 2$ and let $u$ be the function on $\Omega$ which is the constant $0$ on $Q_k$ and $S_k$ for $k<n$, the constant $1$ on $Q_k$ for $k \ge n$, and on $S_k$ for $k>n$, and is linear on $S_n$. Then $\nabla u$ vanishes everywhere, except on $S_n$, where it has constant (pointwise) norm $2^n t_n^{-1}$. The area of $S_n$ is $4^{-n}$, so $$ \int_\Omega \| \nabla u \|^p = 4^{-n} 2^{np} t_n^{-p} = 2^{n(p-2)} t_n^{-p} $$ It is easy to check that $u_\Omega \le 1/2$, so $|u-u_\Omega| \ge 1/2$ on $Q_n$, which also has area $4^{-n}$, implying $$ \int_\Omega |u-u_\Omega|^q \ge 4^{-n} 2^{-q} = 2^{-2n-q} $$ Combining both inequalities, we get $$ \frac{\| u-u_\Omega \|_q}{ \| \nabla u \|_p } \ge \frac{2^{-2n/q-1}}{2^{n(1-2/p)} t_n^{-1}} = 2^{-1+n(2/p-2/q-1)} t_n $$ Here it turns out that we did not even need to keep track of all the constants, we just need to choose any sequence $t_n$ which makes this last term go to infinity (e.g., $t_n = n^n$ works in any case), providing a counterexample.