Let $(M,g)$ be a closed smooth Riemannian manifold. The following is the decomposition part of the Hodge theorem:
Theorem
The canonical map $\mathscr{H}^k(M)\to H^k(M)$ from harmonic $k$ forms into the De Rahm cohomology is an isomorphism.
Let us consider $\Omega^*(M)\otimes \mathbb C$ with the following scalar product:
$$\langle\omega,\eta\rangle := \int_M \omega \wedge *\eta$$
This is a pre-Hilbert space and from the definition $d^*$ is the adjoint of $d$. Since $\Delta = d^\mathstrut d^* + d^*d^\mathstrut=(d^\mathstrut+d^*)^2$, if $(d^\mathstrut + d^*)\omega \neq 0$ then from $$\langle \omega, \Delta \omega \rangle =\langle (d^\mathstrut + d^*) \omega, (d^\mathstrut + d^*)\omega\rangle =\|(d^\mathstrut + d^*)\omega\|^2$$
you get that $\Delta \omega$ also is not zero and $\ker(\Delta)=\ker(d^\mathstrut+d^*)$. Now suppose $\omega \notin \ker(d)$:
$$\langle d\omega , (d^\mathstrut + d^*) \omega \rangle = \langle d \omega, d\omega \rangle + \langle d^2 \omega, \omega \rangle = \|d\omega\|^2$$
And we get $\omega \not\in \ker(\Delta)$. Doing the same with $\ker(d^*)$ you get $\ker(\Delta)\subset\ker(d)\cap\ker(d^*)$. This means that the inclusion defined above is well defined.
On the other hand if $\omega \in \ker(d)\cap\ker(d^*)$ cleary $(d^\mathstrut + d^*)\omega=0$. So a form is harmonic if and only if it lies in the joint kernel of $d$ and $d^*$.
Write $\ker(d)=\text{im}(d)\oplus\text{im}(d)^\perp$, where $\text{im}(d)^\perp$ is the subspace of $\ker(d)$ (not $\Omega^*$) that is orthogonal to $\text{im}(d)$.
If $\omega$ is harmonic then $\omega \in \ker(d^*)$ and $\langle d \eta, \omega \rangle=\langle \eta, d^*\omega \rangle=0$ and $\omega$ lies in $\text{im}(d)^\perp$. On the other hand if $\omega \in \text{im}(d)^\perp$ then $0=\langle d^\mathstrut d^*\omega, \omega \rangle = \langle d^* \omega, d^* \omega\rangle$ and $\omega$ lies in $\ker(d^*)$.
This implies that the space of harmonic forms is equal to $\text{im}(d)^\perp$, which implies the theorem above.
My problem is that this is a very elementary proof, it uses only that $\langle,\rangle$ is positive definite, completeness of the vector space is not needed. Yet in most discussions one sees constant reference to the fact that Hodge theorem is non-trivial and uses involved results about the theory of elliptic operators, but this was not used in this derivation.
Is there an error in the derivation or a missed subtlety? Or does this part of the Hodge theorem (which to me seems the more interesting part) indeed just follow from elementary considerations?
As Daniel Fischer pointed out in the comments, the gap in your proof is the claim that $\operatorname{ker}(d) = \operatorname{im}(d) \oplus \operatorname{im}(d)^\perp$. But it should be noted that simply working in a complete space is not enough to justify this claim; it's necessary to prove that $\operatorname{im}(d)$ is closed. All of the operators $\Delta$, $d$, and $d^*$ are unbounded with respect to the $L^2$ norm, and it's perfectly possible for such an operator to have non-closed image.
The real work in proving the Hodge theorem is proving that $\Delta$ is a Fredholm operator on the $L^2$ completion of $\Omega^*(M)$ (i.e., it has finite-dimensional kernel and closed image). From this it follows rather easily that both $d$ and $d^*$ have closed image, from which you can conclude that $\operatorname{ker}(d) = \operatorname{im}(d) \oplus \operatorname{im}(d)^\perp$. Then you can use elliptic regularity (i.e., if $\Delta \omega$ is smooth, then so is $\omega$) to transfer these results back to $\Omega^*(M)$.