Is $\mathbb{C}^2$ homeomorphic to $\mathbb{C}^2 - (0,0)$ with the Zariski topology? A fellow grad student asked me this, I have been playing for a while but have not come up with anything.  Note that $\mathbb{C}$ is homeomorphic to $\mathbb{C} - \{0\}$ in the Zariski topology - just take any bijection and the closed sets (finite sets) will biject as well.  Concocting a similar thing for the plane is harder though.
I think I can show that the rational plane and the rational plane minus the origin are homeomorphic by enumerating the irreducible curves and using a back and forth argument, but I have not written it all up formally to see if I am missing something yet.
I know the question isn't natural from the point of view of algebraic geometry, because one of the objects isn't even a variety.  I think it is still interesting just to see how weird the zariski topology really is.
 A: $\newcommand\C{\mathbb{C}}$
In the Zariski topology on any quasiprojective variety over $\C$, curves are a distinguished class.  They are minimal elements of the set of infinite closed subsets.  (In fact, by this type of construction, you can define closed subvarieties in general, and recognize the dimension of a closed subvariety.)  Then you can look at fibrations of a quasiprojective variety by curves, in particular fibrations of either $C^2$ or $\C^2 \setminus \{0\}$ by curves.  In $\C^2$, there are many pairs of fibrations with the property that every pair of fibers meets in exactly one point.  For instance, you can take the fibrations by any two families of parallel lines.  The question is delicate because there are many algebraic automorphisms of $\C^2$, and therefore many Zariski self-homeomorphisms.  (In fact the Jacobian conjecture is about algebraic automorphisms of $\C^2$.)
I don't think that this is possible in $\C^2 \setminus \{0\}$.  I'm going to handwave some, but I think that it works.  First of all I think that a fibration by curves has to be an algebraic family, only with its parameterization erased.  Suppose that you have two fibrations $F$ and $G$ of $\C^2 \setminus \{0\}$ such that every pairs of fibers meets at one point.  Then on each side, 0 is in the closure of every fiber.  It's either that or it's in the closure of finitely many fibers.  If it were in the closure of finitely many fibers, you would get two families of curves $F'$ and $G'$ in $\C^2$ such that the intersections leap in isolated places from 1 to 2, which is not possible.  (The intersection cardinality is lower semicontinuous where it is finite.)  On the other hand if every fiber of $F$ approaches 0, then I think that $F$ is a projective family of affine curves, I guess a Riemann sphere of curves.  So then a fiber in $G$, which is not projective, would have an algebraic bijection with the projective parameter space of $F$, which is also not possible.
A: This question was analyzed by  Roger Wiegand  in the context of algebraic surfaces over fields $k$ which are algebraic closures of finite fields. 
[1] R. Wiegand, Homeomorphisms of affine surfaces over a finite field, J. London Math. Soc. (2) 18 (1978), no. 1, 28–32.
and in the followup paper:
[2] R. Wiegand, W. Krauter, Projective surfaces over a finite field. Proc. Amer. Math. Soc. 83 (1981), no. 2, 233–237.
In [1] he proves that the affine plane $A^2_k$ (over $k$) is Zariski-homeomorphic to any open nonempty subset of $A^2_k$ (Corollary 7). In particular, $A^2_k$ is homeomorphic to $A^2_k -\{(0,0)\}$. In particular, Greg's argument (as written) cannot be complete (as zero characteristic is not used anywhere in his sketch). 
After proving the corollary Wiegand asks if the same holds for $k={\mathbb R}$ and ${\mathbb C}$. 
In [2] it is proven (amon other things) that every proper nonempty open subset of $P^2_k$ is homeomorphic to either $P^2_k - \{point\}$, or to the affine plane $A^2_k$. 
