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.
According to your proposed definition, the sequence $1,2,3,\dots$ would be Cauchy in $\mathbb{R}$, witnessed by the sequence of open sets $U_n = (n,\infty)$.
Edit: Let me incorporate some information from the comments to make this a more complete answer.
As another example of why your definition is unsatisfactory: In $\mathbb{R}^2$, any sequence $(a_n,0)$ of points on the $x$-axis is Cauchy, witnessed by the sequence of open sets $\mathbb{R}\times (-1/n,1/n)$.
The fact that completeness and Cauchyness are not topological properties can be formalized by the fact that there are generally many different metrics compatible with a given topology, and these different metrics can induce different notions of completeness and Cauchyness. Looked at a different way, homeomorphisms preserve all topological properties (I would take this to be the definition of a topological property), but homeomorphisms between metric spaces do not necessarily preserve completeness (see the examples in btilly's and Andreas Blass's comments).
On the other hand, the notion of completeness actually lives somewhere in between the world of topological properties and metric properties, in the sense that many different metrics can agree about which sequences are Cauchy. It turns out that they agree when they induce the same uniform structure on the space. And indeed, completeness can be defined purely in terms of the induced uniform structure, so it's really a uniform property.
There is one class of spaces in which topological property and uniform properties coincide: a compact space admits a unique uniformity. So you could call completeness a topological property for compact spaces. But in a rather trivial way, since every compact uniform space is automatically complete.
It could be worthwhile to view compactness as the proper topological version of completeness, i.e. the topological property that comes the closest to agreeing with the uniform/metric property of completeness.
Best Answer
Many of these hold for sequential spaces. These can be defined in a variety of equivalent ways. One simple way that uses no new terminology is that $X$ is sequential iff for each non-closed $A\subseteq X$ there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ converging to a point of $(\operatorname{cl}A)\setminus A$. It turns out that this is also equivalent to the statement that continuity of functions on $X$ is determined by sequences: $X$ is sequential iff for every space $Y$ a function $f:X\to Y$ is continuous iff it preserves convergent sequences, i.e., iff $\langle f(x_n):n\in\Bbb N\rangle$ converges to $f(x)$ in $Y$ whenever $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in $X$.
Sequential compactness and countable compactness are equivalent in sequential space, but unlike the situation in metric spaces, they are not equivalent to compactness: the space of countable ordinals with the linear order topology is first countable, hence sequential, and both countably and sequentially compact, but it is not compact.
If $X$ is second countable (i.e., has a countable base for the topology), then it is compact iff it is sequentially compact, as shown in the answer to this question, but that is more than is needed; for instance, the comments under the question show that they two are equivalent in Lindelöf Hausdorff spaces. (Every second countable space is sequential and Lindelöf, but a sequential Lindelöf space need not be second countable.) The comments also note, with a reference, that these types of compactness are equivalent for the weak topology on Banach spaces, which is a sequential only if the space is finite-dimensional.