It doesn't differ - this is just the subspace topology on $X$.
Given a topological space $A$ and a subset $B\subseteq A$, we can give $B$ the subspace topology.
Let us suppose $B$ is a closed subset of $A$.
Because we have made $B$ into a topological space via the subspace topology, we speak of subsets of $B$ being open or closed.
By the definition of the subspace topology, $C\subseteq B$ is open in $B$ if and only if there exists some $E\subseteq A$, that is open in $A$, such that $E\cap B = C$.
By the definition of "closed", the complement of $E$ in $A$, namely $A\setminus E$, is a closed subset of $A$.
Because $A\setminus E$ and $B$ are both closed subsets of $A$, their intersection $(A\setminus E)\cap B$ is also a closed subset of $A$.
Their intersection is equal to $B\setminus (B\cap E)=B\setminus C$.
Thus, $C\subseteq B$ is open in $B$ if and only if $B\setminus C$ is a closed subset of $A$.
In this case, $A=\mathbb{A}^n$ is affine space with the Zariski topology, $B=X$, and $C=U$.
In the quote from Shafarevich, at the end there appears "$X\setminus U$ is closed", which can be ambiguous when it is said without reference to which topological space it is closed in. However, it is implicitly being taken to mean in the big space, i.e. $A$.
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.
Best Answer
The Zariski topology on $K$, where $K$ is a finite field, is indeed Hausdorff. However, as soon as $K$ is infinite, it is not Hausdorff. This might be interesting, if you are reading a book on algebraic geometry. The Zariski topology is the coarsest topology which satisfies the $T_1$ separation axiom, which means that singleton sets are closed.
So let $K$ be an infinite field. Then The Zariski topology over $K$ is not Hausdorff. Indeed, if we have any two nonempty open sets $U$ and $V$ , then $U^c$ and $V^c$ are finite, so $(U \cap V )^c = U^c \cup V^c$ is finite as well. In particular, $U \cap V$ is nonempty. Thus, open neighborhoods of any two points $x$ and $y$ will intersect.