Is there an intuitive explanation to the fact that the solutions to the time-independent Schrödinger equation form a complete basis? We were always told that the solutions to the time-independent Schrödinger equation form a complete eigenbasis for the space of all functions (all functions?) but I never understood why this is the case. Is there a logical explanation or is it just a weird coincidence?
 A: It is a mathematical theorem that self-adjoint operators in Hilbert space have a complete spectrum.
Note that "self-adjoint" has a special mathematical meaning. Not every Hermitian symmetric operator is self-adjoint. For example, the 1D free Schrodinger Hamiltonian on an open interval without boundary conditions is not self-adjoint. The reason is that we can choose periodic boundary conditions with different phase shifts. Different choices define different self-adjoint operators with different eigenfunctions and eigenvalues. Some potentials do not define self-adjoint Schrodinger Hamiltonians. It can be difficult to prove whether a given potential does so.
A: The "time-independent Schrödinger equation" is just an equation for the eigenvalues and eigenvectors of the Hamiltonian operator on the Hilbert space of states (typically $L^2(\mathbb{R}^3,\mathrm{d}x)$, the "space of wavefunctions")
The spectral theorem tells us that the eigenvectors of any self-adjoint operator form a basis for the space the operator lives on, so as long as your Hamiltonian is self-adjoint, this will hold. (Care has to be taken - Hamiltonians used in physics might only be essentially self-adjoint, or merely Hermitian)
So there is nothing special about the "time-independent Schrödinger equation" - the eigenvectors of other self-adjoint observables also form such bases. No "intuiton" exists for this because it's just a general consequence of the fact/axiom that quantum mechanical observables typically are self-adjoint. You might also pick self-adjoint operators that don't model any physical observable, and you still get this.
