The image of the exterior derivative is closed: Hodge theory If we have an operator $T$ in a Hilbert space $H$ then one can decmpose $H$ as an orthogonal sum $ker(T^*) \oplus \overline{ran(T)}$. This is purely operator theoretic context. On the other hand there is an important theory which is called Hodge theory: one possible application of this theory is to find a special (unique) representative of cohomology class (as a harmonic differential form). My question is the following: 

Does the Hodge theory imply that the range of the exterior derivative is closed?

 A: Sure. Given a degree $d$ differential operator with smooth coefficients $D: \Gamma^\infty(E) \to \Gamma^\infty(F)$ between spaces of smooth sections of some vector bundles, there is a corresponding map $D: L^2_k(E) \to L^2_{k-d}(F)$, given by extension to the closure. And there is also a corresponding map $D^*: L^2_k(E) \to L^2_{k-d}(F)$, the "formal adjoint", which satisfies $\langle Df, g\rangle_{L^2} = \langle f, D^* g\rangle_{L^2}$. 
You are now in a purely functional-analytic context: you have a "closed densely defined unbounded operator" $D: L^2(E) \to L^2(F)$, with natural domain $L^2_d(E)$. The adjoint of this operator in the sense of densely defined operators is a map $D^*: L^2(F) \to L^2(E)$, with natural domain $L^2_d(F)$. And in this context, the formula $\text{Im}(D) = \text{ker}(D^*)^\perp$ as a subset of $L^2(F)$ remains true.
If you want this formula to hold for any Sobolev space, you can bootstrap this to an $L^2$-orthogonal direct sum decomposition of $L^2_k(F)$ formally, so long as $\text{Im}(D)$ is closed. But this is what you were after! So I don't think this strategy is very helpful.
For what remains you need the power of 'elliptic regularity'. I will state this here for the level of generality of "Fredholm complexes", because this is what you are looking at in your post. 
Let $$0 \to \Gamma(E_0) \xrightarrow{D_1} \cdots \xrightarrow{D_n} \Gamma(E_n) \to 0$$ be a complex of differential operators of common degree $d$ over a single manifold $M$, meaning $D_i D_{i-1} = 0$. If this is an elliptic complex, meaning the corresponding linear maps of vector bundles $$0 \to T^*(E_0 \setminus 0) \xrightarrow{\sigma(D_1)} \cdots \xrightarrow{\sigma(D_n)} T^*(E_n \setminus 0) \to 0$$ form, fiberwise, an exact complex. 
This implies that $\Delta_D = \oplus_i (D_i + D_i^*)^2$ is an elliptic operator. In particular, observe that we have the ellieptic regularity theorem: if $\Delta_D \omega \in L^2_{k-d}$, then $\omega \in L^2_{k+d};$ in fact we have a continuous right inverse $F: \text{Im}(\Delta_D) \cap L^2_{k-d} \to L^2_{k+d}$. So if $\Delta_D \omega_n \to \psi \in L^2_k$ for some $k$, then $\Delta_D F \Delta_D \omega_n \to \Delta_D F \psi = \psi$; because $F\Delta_D \omega_n \in L^2_{k+d}$, we see that we have constructed a sequence in $L^2_{k+d}$ so that after applying $\Delta_D$, it converges to $\psi$, as desired.
That is, $\text{Im}(\Delta_D)$ is always a closed subspace (of the appropriate Sobolev space) when $D$ is elliptic; its $L^2$-orthogonal complement is $\text{ker(\Delta_D)}$.
As for the rest, observe that $\text{Im}(D^*D)$ and $\text{Im}(DD^*)$ are $L^2$-orthogonal, and hence never intersect. Because they are further $L^2$ orthogonal with $\text{ker(\Delta_D)}$, we may write eg $DD^* \omega = \Delta_D \psi$ for any $\omega$ and some appropriate $\psi$; thus $DD^* \omega = DD^* \psi + D^* D \psi$, and hence $DD^* (\omega - \psi) = D^* D \psi$; thus $D^*D \psi = 0$. So what we see is we can write any element of $\text{Im}(\Delta_D)$ as a sum of two unique elements of $\text{Im}(D^*D)$ and $\text{Im}(DD^*)$. You can use this fact to argue as above that these subspaces are closed.
From here, I claim that $\text{Im}(D D^* D) = \text{Im}(D^* D) = \text{Im}(D^*)$ and similarly for $\text{Im}(DD^*)$. This is just as straightforward; write $\omega = DD^* \omega_0 + \omega_h + D^*D \omega_1$ where $\omega_h$ is harmonic. Then $D\omega = DD^*D\omega_1$. (You appeal to the regularity results for $\Delta_D$ to ensure that $D^* D \omega_1$ is of the same regularity.) 
Thus we always have the direct sum decomposition $L^2_k(E_i) = \text{Im}(D) \oplus \text{Im}(D^*) \oplus \text{ker}(\Delta_D)$, where all terms are $L^2$-orthogonal, whenever the $D_i$ form an elliptic complex. This is certainly the case for the Hodge complex. It implies the following elliptic regularity statement for Fredholm complexes:

Given a Fredholm complex $(D_i)$ of differential operators of degree $d$ over some closed manifold, if one has $D_i \omega \in L^2_k$, then one may write $\omega = D_{i-1} \omega_0 + \omega_1$, where $\omega_1 \in L^2_{k+d}$.

