In my functional analysis course, my teacher told us that if $\Omega$ is bounded, the two following norms $$u \mapsto \|\nabla u\|_{L^2(\Omega)} = \|u\|_{H^1_0(\Omega)} \quad \text{and} \quad u \mapsto \|u\|_{L^2(\Omega)} + \|\nabla u\|_{L^2(\Omega)} = \|u\|_{H^1(\Omega)}$$ are equivalent on $H^1_0(\Omega)$ thanks to Poincaré inequality. However, the other day I was reading a corollary of the open mapping theorem that states the following:
Let $E, F$ be two Banach spaces and $T$ be linear, continuous and bijective from $E$ to $F$. Then $T^{-1}$ is continuous from $F$ to $E$.
From this we easily deduce that if $\|\cdot\|_1, \|\cdot\|_2$ are two norms on a Banach space $E$ such that there exists $C> 0$ with $$\|x\|_1\le C \|x\|_2, \quad \forall x \in E,$$ then there exists $c > 0$ with $$\|x\|_2\le c \|x\|_1, \quad \forall x \in E,$$ and the two norms are equivalent.
In our case, clearly $$\|u\|_{H^1_0(\Omega)} \le \|u\|_{H^1(\Omega)},$$ on $H^1_0(\Omega)$ whatever is $\Omega$ (no need to be bounded). Therefore, why do we need Poincaré inequality (and hence $\Omega$ bounded) to deduce the equivalence of the norms ? Isn't it straightforward by the corollary above ?
As mentioned in the comments, you would have to show that $H^{1}_{0}(\Omega)$ with this new norm is a Banach space to apply this argument. Note however that it is not necessary that $\Omega$ is bounded for the Poincaré inequality to hold:
If $\Omega \subset \mathbb{R}^n$ and there exists some $j \in \{1,...,n\}$ and $M>0$ such that $\forall x \in \Omega$ we have $|x_j| \le M$ then there is some $C$ such that $$\|u\|_{H^{1}_{0}(\Omega)} \le \|\nabla u\|_{L^2(\Omega)}$$ for all $u \in H^{1}_{0}(\Omega)$ and thus the two norms are equivalent.
A proof can be found here - note that it is sufficient to prove the statement for the dense subset of smooth functions with compact support in $\Omega$.
If $\Omega$ is even bounded then by the Rellich-Kondrachov theorem $H^{1}_{0}(\Omega)$ is compactly embedded in $L^{2}(\Omega)$, which immediately gives equivalency of the two norms.