Why are states rays?
(Answer to OP's 1. and 2.)
One of the fundamental tenets of quantum mechanics is that states of a physical system correspond (not necessarily uniquely - this is what projective spaces in QM are all about!) to vectors in a Hilbert space $\mathcal{H}$, and that the Born rule gives the probability for a system in state $\lvert \psi \rangle$ to be in state $\lvert \phi \rangle$ by
$$ P(\psi,\phi) = \frac{\lvert\langle \psi \vert \phi \rangle \rvert^2}{\lvert \langle \psi \vert \psi \rangle \langle \phi \vert \phi \rangle \rvert}$$
(Note that the habit of talking about normalised state vectors is because then the denominator of the Born rule is simply unity, and the formula is simpler to evaluate. This is all there is to normalisation.)
Now, for any $c \in \mathbb{C} - \{0\}$, $P(c\psi,\phi) = P(\psi,c\phi) = P(\psi,\phi)$, as may be easily checked. Therefore, especially $P(\psi,\psi) = P(\psi,c\psi) = 1$ holds, and hence $c\lvert \psi \rangle$ is the same states as $\lvert \psi \rangle$, since that is what having probability 1 to be in a state means.
A ray is now the set of all vectors describing the same state by this logic - it is just the one-dimensional subspace spanned by any of them: For $\lvert \psi \rangle$, the associated ray is the set
$$ R_\psi := \{\lvert \phi \rangle \in \mathcal{H} \vert \exists c \in\mathbb{C}: \lvert \phi \rangle = c\lvert \psi \rangle \}$$
Any member of this set will yield the same results when we use it in the Born rule, hence they are physically indistiguishable.
Why are phases still relevant?
(Answer to OP's 3.)
For a single state, a phase $\mathrm{e}^{\mathrm{i}\alpha},\alpha \in \mathbb{R}$ has therefore no effect on the system, it stays the same. Observe, though, that "phases" are essentially the dynamics of the system, since the Schrödinger equation tells you that every energy eigenstate $\lvert E_i \rangle$ evolves with the phase $\mathrm{e}^{\mathrm{i}E_i t}$.
Obviously, this means energy eigenstates don't change, which is why they are called stationary states. The picture changes when we have sums of such states, though: $\lvert E_1 \rangle + \lvert E_2 \rangle$ will, if $E_1 \neq E_2$, evolve differently from an overall multiplication with a complex phase (or even number), and hence leave its ray in the course of the dynamics! It is worthwhile to convince yourself that the evolution does not depend on the representant of the ray we chose: For any non-zero complex $c$, $c \cdot (\lvert E_1 \rangle + \lvert E_2 \rangle)$ will visit exactly the same rays at exactly the same times as any other multiple, again showing that rays are the proper notion of state.
The projective space is the space of rays
(Answer to OP's 4. and 5. as well as some further remarks)
After noting, again and again, that the physically relevant entities are the rays, and not the vectors themselves, one is naturally led to the idea of considering the space of rays. Fortunately, it is easy to construct: "Belonging to a ray" is an equivalence relation on the Hilbert space, and hence can be divided out in the sense that we simply say two vectors are the same object in the space of rays if they lie in the same ray - the rays are the equivalence classes. Formally, we set up the relation
$$ \psi \sim \phi \Leftrightarrow \psi \in R_\phi$$
and define the space of rays or projective Hilbert space to be
$$ \mathcal{P}(\mathcal{H}) := (\mathcal{H} - \{0\}) / \sim$$
This has nothing to do with the Gram-Schmidt way of finding a new basis for a vector space! This isn't even a vector space anymore! (Note that, in particular, it has no zero) The nice thing is, though, that we can now be sure that every element of this space represents a distinct state, since every element is actually a different ray.1
(Side note (see also orbifold's answer): A direct, and important, consequence is that we need to revisit our notion of what kinds of representations we seek for symmetry groups - initially, on the Hilbert space, we would have sought unitary representations, since we want to preserve the vector structure of the space as well as the inner product structure (since the Born rule relies on it). Now, we know it is enough to seek projective representations, which are, for many Lie groups, in bijection to the linear representations of their universal cover, which is how, quantumly, $\mathrm{SU}(2)$ as the "spin group" arises from the classical rotation group $\mathrm{SO}(3)$.)
OP's fifth question
When one limits the Hilbert space to that of a certain observable of the system at hand, e.g. momentum or spin space (in order to measure the momentum and spin of a system respectively), does that mean we're talking about projective spaces already? (e.g. is the spin space spanned by up |↑⟩ and down |↓⟩ spins states of a system referred to as projective spin Hilbert space?)
is not very well posed, but strikes at the heart of what the projectivization does for us: When we talk of "momentum space" $\mathcal{H}_p$ and "spin space" $\mathcal{H}_s$, it is implicitly understood that the "total space" is the tensor product $\mathcal{H}_p \otimes \mathcal{H}_s$. That the total/combined space is the tensor product and not the ordinary product follows from the fact that the categorial notion of a product (let's call it $\times_\text{cat}$) for projective spaces is
$$ \mathcal{P}(\mathcal{H}_1) \times_\text{cat} \mathcal{P}(\mathcal{H}_2) = \mathcal{P}(\mathcal{H}_1\otimes\mathcal{H}_2)$$
For motivations why this is a sensible notion of product to consider, see some other questions/answers (e.g. this answer of mine or this question and its answers).
Let us stress again that the projective space is not a vector space, and hence not "spanned" by anything, as the fifth question seems to think.
1The inquiring reader may protest, and rightly so: If our description of the system on the Hilbert space has an additional gauge symmetry, it will occur that there are distinct rays representing the same physical state, but this shall not concern us here.
Best Answer
One problem with this idea is normalization:
$$\int_{\mathbb R} \psi^* (x) \psi(x)~ dx = 1$$
You are integrating over infinite space. If $\psi$ has a minimum non-zero value, $\psi$ must be $0$ everywhere except a finite volume.
Now switch to the momentum basis. Because $\psi$ has bounded support, the Fourier Transform of it cannot have. To be normalizable, the tails would have to have infinitesimal values. So you cannot have discrete values in momentum space. Does this fit your theory?
Another problem is that wave functions are continuous. If there are only a discrete set of values, you would have discontinuous functions.
Unless you are talking about a space with holes in it? Constant values in distinct regions?
Given
$$\hat p \psi(x) = -i\hbar \frac{\partial\psi}{\partial x}$$
a $\psi$ that was constant, except where interrupted by discontinuities would correspond to $\hat p = 0$ except where it is undefined or perhaps has infinite spikes.
Likewise
$$-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} = E \psi$$
would lead to $E = 0$ except perhaps at the discontinuities.
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