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I'm not claiming Sage is bug-free, but a lot of those bugs aren't mathematical errors -- there are plenty of compilation issues, documentation problems, etc., not to mention nearly 700 tickets classified as "enhancement" rather than "defect" -- so claiming 1582 known bugs is a little misleading.
Let p/q be a close approximation to 2^(1/3), say p/q = 2^(1/3) + e with |e| < 1/(sqrt(5)*q^2); there are infinitely many of these by Hurwitz's theorem. Letting u=v=q, we have (u^3+v^3)^(1/3) = 2^(1/3) * q = p - qe, so the error qe is at most 1/(sqrt(5)*q) in magnitude, and as q goes to infinity the error goes to 0.
It's even easier to use EZproxy: go to libraries.mit.edu/about/faqs/… and add the bookmarklet to your browser toolbar now. (Harvard has it too but I don't know the URL offhand.)
R^3 doesn't have any special properties in that sense. The issue is that curves only have codimension 1 in R^2, so they can separate the plane into multiple components, whereas they can't separate R^3. You could probably formulate a similar problem about using surfaces with boundary to connect 1-manifolds (circles or line segments) in R^3 and get a finiteness result for that if you really wanted.
Then you just realize that det(tI-A) evaluated at A is some matrix whose entries are monstrously complicated polynomials in the n^2 entries of the matrix A, and since they're identically 0 on C^{n^2} each of those entries must be the zero polynomial; thus the theorem holds over any commutative ring as well.
For d=2, every closed orientable manifold is a sphere, disk, annulus, torus, or something hyperbolic -- the sphere is the only one of these which even has nontrivial homotopy groups beyond \pi_1. (For d >= 3 this should be much harder and I don't know what to tell you.)
Actually, some more interesting "finite axioms of choice" have been defined: for each n > 0, the axiom [n] states that every collection of n-element sets admits a choice function. Then it might be natural to ask which collections of axioms [m_i] imply some [n]; e.g. Tarski showed that [2] and [4] are equivalent. See Conway, "Effective implications between the 'finite' choice axioms," MR0360275, which discusses this in detail and works out all the solutions up to n=64.
