Let $\Omega \subset \mathbb{R}^{n}$ be open and bounded. For any $u \in H^{1}(\Omega)$, define $$ \|u\|:=\|u\|_{L^{2}(\Omega)}+\|\nabla u\|_{L^{2}(\Omega)} . $$
How can I show that there exist $C_{1}, C_{2}>0$ (independent of $u$) such that $$ C_{1}\|u\| \leq\|u\|_{H^{1}(\Omega)} \leq C_{2}\|u\| $$
My attempt:
Because of the Poincaré inequality, there is a constant $C$ with $$ \|u\|_{L^{2}(\Omega)} \leq C\|\nabla u\|_{L^{2}(\Omega)} \quad \forall u \in H_{0}^{1}(\Omega) $$ It follows for $u \in H_{0}^{1}(\Omega) \subset H^{1}(\Omega)$ that $$ \begin{aligned} \|u\|_{H^{1}(\Omega)} &=\left(\|u\|_{L^{2}(\Omega)}^{2}+\|\nabla u\|_{L^{2}(\Omega)}^{2}\right)^{1 / 2} \leq\left(C\|\nabla u\|_{L^{2}(\Omega)}^{2}+\|\nabla u\|_{L^{2}(\Omega)}^{2}\right)^{1 / 2} \\ & \leq C\|\nabla u\|_{L^{2}(\Omega)} \leq C\|u||_{H^{1}(\Omega)} \end{aligned} $$
Is this correct? How do I show the full inequality?
Even if your approach were correct, you would be proving the inequality only on a subspace of the Sobolev space, hence not getting the correct result.
If I understood well you are proving the equivalence of the following two norms in $H^1$:
$$ \|u\|:=\|u\|_{L^{2}(\Omega)}+\|\nabla u\|_{L^{2}(\Omega)} . $$
$$ \|u\|_{H^{1}(\Omega)} =\left(\|u\|_{L^{2}(\Omega)}^{2}+\|\nabla u\|_{L^{2}(\Omega)}^{2}\right)^{1 / 2} $$
I would not use Poincaré inequality here, but only the same strategy you use to prove equivalence between euclidean norm and taxi-cab norm (or of $p$ norms in finite dimensional spaces) (let me give you a hint below)