inverse laplace operator bounded

Given $\Omega \subset \mathbb R^d $ open and bounded, we define the operator $T:L^2(\Omega) \to H_0^1 (\Omega)$ as $T(f)=u$ where $u \in H_0^1(\Omega)$ is the unique weak solution to the Poisson equation: $-\Delta u=f $ in $\Omega$ and $u=0$ on $\partial \Omega$, meaning $\int_\Omega \nabla u\cdot \nabla v dx = \int_\Omega fvdx$ for all $v\in H_0^1(\Omega)$.

Then, we define an inner product on $H_0^1(\Omega)$: $$<f,g>_{H^{-1}}:=\int_\Omega fT(g)dx. $$

I need to show, that the linear operator $T$ is bounded and that the space $(H_0^1(\Omega), \|\cdot\|_{H^{-1}})$ is not complete. I've tried using the Poincare and Holder inequalities to show the boundedness of $T$. I get for $u=T(f)$:$$ \|u\|_{H_0^1}^2=\|u\|_2^2+\int _\Omega |\nabla u|^2dx\leq (c+1)\int_\Omega |\nabla u|^2dx=(c+1)\int _\Omega fudx \leq (c+1)\|f\|_2\|u\|_2 $$ but I don't know how to proceed from here. Is there something obvious I am missing?

To show that the space is not complete, do I explicitely find a non-convergent Cauchy sequence or do I show that our norm here is equivalent to one with which our space is known to be non-complete?

Any help is greatly appreciated. Thanks!

Here's the answer for showing $T$ is bounded. You almost had it actually. Let me use $Tf$ instead of $u$. From what you had, \begin{align*} \int_\Omega |\nabla(Tf)|^2\, dx = \int_\Omega fTf\, dx & \le \|f\|_{L^2(\Omega)}\|Tf\|_{L^2(\Omega)} \\ & \le C\|f\|_{L^2(\Omega)}\|\nabla(Tf)\|_{L^2(\Omega)}, \end{align*} where we apply the Poincaré inequality (with constant $C$) one more time. This gives $\|\nabla(Tf)\|_{L^2(\Omega)}\le C\|f\|_{L^2(\Omega)}$ and so $$ \|Tf\|_{H_0^1(\Omega)}\le \sqrt{C^2 + 1}\|\nabla Tf\|_{L^2(\Omega)}\le C\sqrt{C^2 + 1}\|f\|_{L^2(\Omega)}. $$