Your solution is valid. It has zero kinetic energy. It doesn't necessarily have zero energy. It can have any potential energy you'd like. Just because your particle is "freely moving," that doesn't mean the potential is zero. You could have $V(x)=k$ for any constant $k$. The value of $k$ is not observable and has no physical significance.
In general there is no special significance to having zero energy in a solution to the Schrodinger equation. Any solution can be defined to have zero energy simply by changing the potential appropriately like $V\rightarrow V+c$, where $c$ is some constant.
A realistic example involving zero kinetic energy and a constant wavefunction would be some particle-rotor models of nuclei, in which the deformed (prolate) nucleus (rotor) has some orientation in space, specified by one or two angular coordinates. If the odd particle has some component $K$ of its angular momentum along the symmetry axis, you get a rotational band with energies proportional to $J(J+1)$, starting with a ground state at $J=K$. In the ground state for the $K=0$ case, the rotor has zero kinetic energy, and its wavefunction is a constant as a function of the angular coordinates.
I'm answering only part of your question, please bear in mind that some of the topics that you pointed out are doubts I have myself.
Concerning the comment from Feynman, I usually don't like the authority arguments. If your only reason to believe something is that someone said it, it's not a good reason. Just as some examples, Newton believed that light had no wave-like behaviour, which is simply wrong.
About the variational formulation of Schrödinger Equation. If by variational you mean, having a Lagrangian, the following Lagrangian does the job:
$
\mathcal L = \frac{i\hbar}{2}\left(\psi^*\partial_t \psi - \psi \partial_t \psi^*\right) - \frac{\hbar^2}{2m}\left(\nabla \psi\right)\cdot \left(\nabla \psi^*\right) - V\psi^*\psi
$
With the assumption that you have a complex classical scalar field, $\psi$, and that you can calculate the Euler-Lagrange equation separately for $\psi$ and $\psi^*$. As with anything using the extremum action principle, you really have to guess which your Lagrangian is based in Symmetry principles + some guideline for your problem, here is no different.
I don' t know how did Schrödinger made the original derivation, but if you make a slight change of variables in the above lagrangian, you get very close to what you have written above.
The trick is writing $\psi = \sqrt n e^{iS/\hbar}$, where $n$ is the probability density and $S$ is, essentially the phase of the wave-function. If you have trouble with the derivation, I can help you latter.
Again about Feynman's quote. It's possible to arrive at Schrödinger equation without passing through an arbitrary heuristic procedure, that way is developed by Ballentine's book. You still have to postulate where you live, if you consider that arbitrary or not, it is completely up to you.
The other point that it is possible to consistently and rigorously construct a quantum theory from any classical mechanical theory, using quantization by deformation. That's why I think it's a bit false that "It's not possible to derive it from anything you know. It came out of the mind of Schrödinger", as Feynman, and many other great names, said.
Best Answer
It is extremely misleading to call the argument you referred to as a "derivation" of the Schrödinger equation $-$ to put it simply, it is no such thing, and there is no such thing. The Schrödinger equation is one of the fundamental postulates of quantum mechanics and we know of no deeper justification for it.
The argument you've sumarized is best described as a justification of why the Schrödinger equation is, hopefully, not too much of an unreasonable thing to try. It is an argument that makes a bunch of leaps of logic, some of which you've already identified, and it cannot be fixed. That doesn't make the argument any less useful, but to the degree that you find logical flaws in it, they are mostly really there.