Bound on fermions in a finite volume? The Pauli Exclusion Principle says that two or more identical fermions cannot occupy the same quantum state within a quantum system simultaneously. However, I'm wondering if we could potentially pack an infinite number of fermions within a finite volume? 
Although it is wrong to think of these as point particles... Intuitively my idea is: Suppose some fermion is at some point $(x,y,z)$ then what prevents another fermion occupying the position $(x+\varepsilon,y+\varepsilon,z+\varepsilon)$ where $\varepsilon$ is very small.
Case 1: What's keeping $\varepsilon$ to vary across all values of $[0,1]$ so that there are literally an uncountable number of particles in a ball of radius $1$ centered at $(x,y,z)$? 
Case 2: What's keeping $\varepsilon$ from varying as $\varepsilon=\frac{1}{n}$ for all $n\in \mathbb{N}$ so that there are a countable number of fermions in a ball of radius $1$ centered at $(x,y,z)$?
Case 3: Is there any way to fit a countable number of fermions within a ball of finite radius?
 A: First, a point of clarification.

Although it is wrong to think of these as point particles...

This is critical.  The state of a particle in a box cannot be described by a set of position coordinates.  
Consider a 1-D box of length $L$ with two indistinguishable fermions inside it.  The wavefunction of the two-particle system  $f(x_1,x_2)$ must obey the boundary conditions $f(0,x_2)=f(L,x_2)=f(x_1,0)=f(x_1,L)=0$ (I'm going to ignore normalization for simplicity).
You can interpret $x_1$ and $x_2$ as the "position" coordinates for the two fermions, in analogy with the single particle in a box problem from elementary quantum mechanics.  The (fermionic) indistinguishability condition places an additional constraint on the state - namely that $f(x_1,x_2)=-f(x_2,x_1)$.
What kind of functions obey these conditions?  Well, the single particle energy eigenstates of the 1D particle in a box take the form
$$f_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$$
so one guess would be the function
$$\frac{2}{L} \sin\left(\frac{\pi x_1}{L}\right) \sin\left(\frac{2\pi x_2}{L}\right)$$
This obeys the boundary conditions, but does not obey the antisymmetry condition - the correct antisymmetrized version is
$$f_{12}(x_1,x_2) = \frac{2}{L}\left[ \sin\left(\frac{\pi x_1}{L}\right) \sin\left(\frac{2\pi x_2}{L}\right) - \sin\left(\frac{\pi x_2}{L}\right) \sin\left(\frac{2\pi x_1}{L}\right)\right]$$
so what does this state (or rather, the probability density $|f(x_1,x_2)|^2$) look like?

Remember that $|f(x_1,x_2)|^2| dx_1 dx_2$ is the probability of measuring one fermion in the interval $[x_1, x_1+dx_1]$ and one in the interval $[x_2,x_2+dx_2]$.  As expected, along the line $x_1=x_2$, the wavefunction is equal to zero - physically, this means that the probability of measuring two indistinguishable fermions in the same infinitesimal interval in the 1D box is equal to zero.  
Furthermore, if we fix $x_1$ at some position (say, $x_1=0.25$) then we see that the probability of finding the other particle near that position is vanishingly small, and has a maximum on the other side of the box (around $x_2=0.75$).
If we use more complex states (say, $n=2$ and $n=4$), we get more complicated probability density functions

but we will always have that $x_1=x_2 \implies f(x_1,x_2)=0$ due to the asymmetry condition on fermionic wavefunctions.

Now to your questions.

What's keeping $\epsilon$ to vary across all values of $[0,1]$ so that there are literally an uncountable number of particles in a ball of radius $1$ centered at $(x,y,z)$?

I don't really know how one would even describe an uncountable number of particles, so let's put a pin in this.  Maybe somebody else can provide a satisfactory answer.

What's keeping $\epsilon$ from varying as $\epsilon = \frac{1}{n}$ for all $n\in\mathbb N$ so that there are a countable number of fermions in a ball of radius 1 centered at $(x,y,z)$?

Nothing.
A 2-particle fermionic wavefunction can be any function $f(x_1,x_2)$ which obeys the boundary conditions and is antisymmetric.  An N-particle fermionic wave function can be any function $f(x_1,x_2,\ldots,x_N)$ which obeys the boundary conditions and is antisymmetric in all $N$ of its entries.
If you consider your single particle wavefunctions to be e.g. sharply peaked gaussians, then there's nothing keeping you from adding as many of them as you'd like to your box - as long as the total wavefunction is antisymmetric.  Given any desired collection of $N$ single-particle wavefunctions, you can use the Slater determinant to find the appropriately antisymmetrized combination.
Of course, that antisymmetry will mean that the probability of measuring any two particles in the same infinitesimal interval is equal to zero.

Is there any way to fit a countable number of fermions within a ball of finite radius?

Sure.  If we consider only energy eigenstates (which we don't need to do, but we certainly can), then you should be able to see that although we can't have two particles in the same energy eigenstate, we can have one particle in state $1$, one in state $2$, one in state $3$, and so on.  We can add an arbitrarily large number of particles to our box, as long as we pay the price that each particle we add will need to have a higher energy than the one before it.  This is the concept underlying the Fermi level, a crucial idea in solid state physics.
A: No, there is no such bound. The easiest way to see this is based on units. Given $m$ and $h$, there is no way to form a quantity with units of density.
In the context of general relativity, there is nothing about the underlying quantum mechanics of a fermionic matter field that would cause an equation of state that would prevent strong curvature singularities. For a description of what a strong curvature singularity is, see 
Rudniki et al., "Generalized Strong Curvature Singularities and Cosmic Censorship," https://arxiv.org/abs/gr-qc/0203063
A: The short answer is YES, because in general there will be an infinite number of states available within any finite volume. If you make the number of particles within the volume arbitrarily large, then the highest occupied energy level will also have to be arbitrarily large, but this is not a fundamental problem. This is the basic idea behind the Fermi gas, which is the name for the situation you are describing.
