Is there a way of working with the Zariski topology in terms of convergence/limits? As someone who is very fond of analysis, I feel most comfortable working in topological spaces via the notion of convergence of sequences (or nets, in infinite-dimensional Banach spaces, etc.). In every metric space (or first-countable topological space) anything you can say about the topology can be said in terms of convergent sequences (though perhaps less elegantly). In arbitrary topological spaces, one can use nets instead. However, if the space is not Hausdorff then nets don't have unique limits, which makes it quite difficult to work with them the way one is used to working with sequences (of real numbers, say). This is the case with the Zariski topology on $\mathbb{C}^n$ (I have no urgent desire to consider more general algebraic varieties). Is there some convenient way to express the Zariski topology (that is: its closed sets, the continuous functions it allows,...) in terms of some "convergence structure" which satisfies properties similar to those we have in any sequential/metric/first-countable space?
 A: The Zariski topology is not "really" a topology in the familiar analytic sense. It precisely encodes the fact that if you have polynomials vanishing on a bunch of points, then they must also vanish on a bunch of other points. For example, if a real polynomial in one variable vanishes on $\mathbb{Z}$, it also vanishes identically. Since the Zariski closure of a subspace can often be much larger than the subspace itself, I don't see a good way to shoehorn in analytic intuition where it doesn't belong. The fact that the Zariski topology is not generally Hausdorff is not some peculiarity to be ignored; it is a fundamental and basic fact about polynomials. 
A: The problem I see is that, since the Zariski topology is non-Hausdorff, limits of sequences are not unique. For example, consider the affine line $\mathbb{A}^1$ over an algebraically closed ground field $k$. Let $x_n$ be any sequence of distinct elements in $k$, so we can think of $x_n$ as a sequence of points of $\mathbb{A}^1$. Then, for any point $y$ in $\mathbb{A}^1$, and any open set $U$ containing $y$, we have $x_n \in U$ for $n$ sufficiently large. So $x_n$ approaches every point $y$.
It can often be interesting to talk about the collection of limit points of a sequence in the Zariski topology. However, there is essentially never a unique limit. In my opinion, this would make it very hard to think about the Zariski topology in terms of sequences. 
A: I don't think that there is a useful way to do what you ask (namely to work with
limits/convergence); as other answers have explained, the non-Hausdorff nature of the Zariski topology (among other things) makes this difficult.
On the other hand, most basic lemmas in topology/analysis which can be proved via
a consideration of convergent sequences can normally also be proved via arguments with open sets instead, and so your intuition for the topology of metric spaces can to some extent be carried over, if you are willing to make these sort of translations.  At some point (if you practice), and with a bit of luck, the translation will become automatic (or at least close to automatic).
(Although you may
think of non-Hausdorffness as a serious pathology that invalidates what I've just said, in the end it's less serious psychologically than it seems at first
--- at least in my experience.)
Speaking for myself, I certainly regard the Zariski topology as a topology, just like any other (meaning that I don't think of it as some other thing which happens to satisfy the axioms of a topology; I think of it as a topology in a genuine sense).  It is just that it doesn't allow many closed sets: only those which can be cut out as the zero locus of a polynomial. 
So a good way to practice thinking about the Zariski topology is to more generally practice thinking about topologies in terms of what kinds of closed sets are allowed, or, more precisely, what kinds of functions are allowed to cut out closed sets as their zero loci.  
Thinking in terms of functions is a way of bridging the analytic intuition that you seem to like and the general topological formalism that underlies the Zariski topology.  What I mean is: in standard real analysis, if you have a continuous function on $\mathbb R^n$ (or a subset thereof), its zero locus is closed.  One way to think about this is via sequences (this is a way that you
seem to like): if $f(x_n) = 0$ for each member of a convergent sequence,
then $f(\lim_n x_n) = 0$ too, as long as $f$ is continuous. 
Now, when working with the Zariski topology, you have to throw away the argument with sequences, but you can still keep the consequence: the zero locus of a "continuous" function is closed.  The key point is that now the only functions that you are allowed to think of as being continuous are polynomials.  This may take some getting used to, but is not so bad (after all, polynomials are continuous in the usual topology on $\mathbb C^n$ as well!).  

Summary: It doesn't seem possible to work rigorously with a sequence/convergence point of view, but (a) it is not so misleading for very basic topological facts; and (b) another analytic view-point that is very helpful is to think about the topology in terms of its closed sets being zero-loci of continuous functions --- you just have to restrict the functions that you call "continuous" to be polynomials.
A: Getting a handle on the Zariski topology can be very tough and in my experience I have found that it is easier to learn how to use it then to try to picture it. At least part of the problem stems from the fact that open sets in a variety or scheme are huge and the topology is almost never Hausdorff.
Of course there are many objects in Algebraic geometry that do have nice pictures. A very small list (in no particular order) is given below. Understanding the pictures associated to these vocab words along with the corresponding ideas in algebra may help give you a better handle on the Zariski topology. I recommend Eisenbud and Harris, Geometry of Schemes as a reference along with Ravi Vakils Math 216 notes.
Fat points, Fuzzy points, closed subschemes, codimension one sets, stalks, irreducible sets, singular points...
A: Here are some results which will allow you do decide for yourself how good/bad are schemes as compared to your favorite topological spaces.
For an  affine scheme $X=Spec(R)$ we have:
$$X\; is \; \text T_2 \text{ (=Hausdorff)} \iff X \;\text {is} \; T_1 (=\text {closed points})\iff R \; \text {has dimension 0}$$
Examples of rings of dimension 0 (=rings in which all prime ideals are maximal) :
Trivial examples of zero dimensional rings are finite products of fields or rings of the form $k[T]/(T^n)$ over the field $k$.
The richest source of non trivial zero-dimensional rings are  the von Neumann regular rings.
They are the rings for which the following holds:   for all $ r\in R$  there  exists $a\in R$  such that $r=ar^2$. (These rings are called absolutely flat by Bourbaki and his groupies because every module on such a ring is flat.)
Any infinite product of fields is von Neumann regular  and the spectrum of such a product is thus Hausdorff. But very weird: Mumford refers to these spectra as "outrageous spaces" and speaks of  "far-out mysteries" (in his Red Book, page 140).For example if you take denumerably many copies of a field, the spectrum of their product will be homeomorphic to the Stone-Čech compactification of $\mathbb N$.
Edit (later)
In defense of that poor Zariski topology, I would like to make the point that it satisfies  a very strong separation property !
Recall that in Topology a Hausdorff space $X$ is said to be normal if  any two disjoint closed subsets have disjoint neighborhoods. Urysohn famously proved that these spaces have the property  that given two disjoint closed subsets $A,B \subset X$ there is a real-valued function $f: X\to \mathbb R$  with $f|A=0$ and $f|B=1$.
 Well, for affine schemes you can do better than that and find a global regular function on $X$ that will simultaneously  extend any regular functions given on
$A$ and $B$:
Urysohn for affine schemes Let  $X=Spec(R)$ be an affine scheme and $A=V(I)$, $B=V(J)$ two disjoint closed subschemes.Then for any $f_A\in \Gamma (A, \mathcal O_A), f_B\in \Gamma (B, \mathcal O_B) $ there exists $f\in \Gamma (X, \mathcal O_X)$ with $f|_A=f_A$ and $f|_B=f_B$
Is that a brand-new result that hasn't made its way into textbooks yet? Not at all: it is the third century A.D. Chinese Remainder Theorem! You see, the disjointness of $A=V(I)$ and  $B=V(J)$ translates into the comaximality $I+J=R$
and the surjectivity of $R\to R/I \times R/J$ then proves Urysohn for affine schemes.
