Let $\Omega \subset \mathbb{R^n}$ be open. Let $H = H_0^m(\Omega) = \overline{\mathcal{D(\Omega)}}$, with the usual Sobolev-norm.
For $f \in L^2(\Omega), g \in H$, $g \mapsto \int_{\Omega}(fg)dx$ is an element of $H'$.
Let $\|f\|_{-m}$ be the norm of this functional.
Show: Completion of $(L^2(\Omega), \|.\|_{-m}) \cong H'$
Here's my attempt:
By definition one can say $(L^2(\Omega), \|.\|_{-m}) \subset H'$ is subspace of $H'$. Thus $\overline{(L^2(\Omega), \|.\|_{-m})}$ is a closed subspace of $H'$.
$H$ is a Hilbert-space, so is $H'$, and thus $\overline{(L^2(\Omega), \|.\|_{-m})}$.
Final step:
$H$ is a separable and infinitely dimensional. Because $H \cong H'$, $H'$ is separable and infinitely dimensional.
Because $\overline{(L^2(\Omega), \|.\|_{-m})} \subset H'$, it is also separable and clearly infinitely dimensional. Because all seperabel, infinitely dimensional Hilbert-spaces are isomorphic to each other $H' \cong \overline{(L^2(\Omega),\|.\|_{-m})}$ is true.
My question is: Is it legit to identify the completion of $(L^2(\Omega), \|.\|_{-m})$ with its closure as a subset of $H'$? The textbook definition of a completion is to look at all Cauchy-Sequences and so on, which is not exactly what i'm doing.
Your proof is correct if you just want to show that $H'=(H^m_0(\Omega))'$ and $\overline{(L^2(\Omega),\lVert \cdot \rVert_{-m})}$ are isometrically isomorphic. For this, it suffices to observe that both are separable and infinite dimensional Hilbert spaces, after which you can appeal to a general result.
However, I suspect what you really want to show is that the map $$ T : (L^2(\Omega),\lVert \cdot \rVert_{-m}) \rightarrow H' \text{ sending } f \mapsto \left(g \mapsto \int_{\Omega} fg \,\mathrm{d}x\right)$$ extends to an isometric isomorphism. Here by passing to the completion, we obtain a unique extension $$ \widetilde T : \overline{(L^2(\Omega),\lVert \cdot \rVert_{-m})} \rightarrow H' $$ which agrees with $T$ on $L^2(\Omega)$. Assuming $T$ is an isometry (which you need to show), it is not difficult to see that $\widetilde T$ inherits this property. Thus it remains to show that $\widetilde T$ is surjective, which is missing in your proof.
This distinction is subtle but important; with function spaces the duality is often identified using the pairing $\langle f, g \rangle = \int_{\Omega} f g \,\mathrm{d}x$, and it is important in applications that they aren't merely isomorphic. This is because for instance $H^m_0(\Omega)$ and $L^2(\Omega)$ are also isomorphic as Hilbert spaces, but you don't want to think of these spaces as the same when $m > 0$.
Also to answer your question at the end, one generally identifies the completion of $L^2(\Omega)$ with respect to the norm $\lVert \cdot \rVert_{-m}$ as a subset of $H'$ via the map $\widetilde T$, even though they aren't strictly equal as you say. In general you want to think of elements of these function spaces as functions, or at least distributions (in $\mathcal D'(\Omega)$ or $\mathcal S'(\Omega)$). For an extended discussion on this, I will refer you to a related answer of mine.
To show surjectivity of $\widetilde T$, this depends on how you define your Sobolev norm and also $\lVert \cdot \rVert_{-m}$. One way is to show this, which is motivated by the linked answer above, is to show that for $L \in H'$, there exists $\{f_{\alpha}\}_{\lvert \alpha \rvert \leq m}$ such that $$ L(g) =\sum_{\lvert \alpha \rvert \leq m} (-1)^{\lvert \alpha\rvert}\int_{\Omega} f_{\alpha} D^{\alpha}g \,\mathrm{d} x$$ for each $g \in H$. Then we can approximate each $f_{\alpha}$ in $L^2(\Omega)$ by $f_{k,\alpha} \in C^{\infty}_c(\Omega)$ and show that $$ L_k(g) =\sum_{\lvert \alpha \rvert \leq m} (-1)^{\lvert \alpha\rvert}\int_{\Omega} f_{k,\alpha} D^{\alpha}g \,\mathrm{d} x$$ defines a sequence of elements converging to $L$ in $H'$. Now one can check that $$ L_k = T(f_k) \text{ where } f_k = \sum_{\lvert \alpha \rvert \leq m} D^{\alpha} f_{k,\alpha} \in C^{\infty}_c(\Omega) \hookrightarrow L^2(\Omega),$$ so it follows that the image of $T$ is dense in $H'$. This implies that $\widetilde T$ must be surjective, as required.