Edit, 7/10/11: The idea below was recently also discussed in this blog post by Michael O'Connor.
This was thoroughly discussed on MathOverflow. The answers that I found most convincing can be summarized as follows: open sets axiomatize the notion of a semidecidable property.
That is, open sets axiomatize the notion of a condition whose truth can be verified in finite time (but whose falsehood cannot necessarily be verified in finite time). A continuous function $f : A \to B$ between two spaces is a function such that the preimage of any semidecidable subset is semidecidable, hence is a computable function in the sense that a decision procedure to verify whether $f(a) \in U \subset B$ in finite time gives a decision procedure to verify whether $a \in V \subset A$ in finite time.
To really make sense of what I just said above you should think of $A$ as the set of possible conditions of some system, $f$ as a measurement of some property of the system, and $B$ as the set of possible values of the property you're measuring. Then $f$ is computable precisely when information about $f(a)$ allows us to deduce information about $a$. In some sense it is the central premise of the scientific method that this is possible.
Note that the above description really brings out the special role of the Sierpinski space $\mathbb{S}$. Indeed, a subset of a topological space $X$ is open precisely when the indicator function $X \to \mathbb{S}$ is continuous.
Onto the axioms:
- The union of an arbitrary collection of open sets is open because any decision procedures to verify conditions $U_i, i \in I$ in finite time can be run in parallel to verify condition $\bigcup U_i$ in finite time.
- The intersection of a finite collection, but not necessarily an infinite collection, of open sets is open because a finite number of decision procedures to verify conditions $U_1, ... U_n$ in finite time can be run in parallel to verify condition $\bigcap U_i$ in finite time, but this is not necessarily true of an infinite number of decision procedures, which may take an unbounded amount of time to all complete.
- The empty set and the entire space are open because both of these conditions can be verified in zero time.
Finally, note that it is intuitively possible to verify whether a point in a metric space lies in an open ball in finite time by showing that it lies in an even smaller open ball (which can be done using finite-precision computations), but it is not necessarily possible to do the same for a closed ball because the point may lie on the boundary and we cannot make arbitrarily precise measurements in finite time.
In the particular case of the Zariski topology, it is always possible to verify in finite time whether a polynomial is nonzero at a point by computing with sufficient precision, but without additional information it is not necessarily possible to verify in finite time whether a polynomial is zero at a point.
The word "standard" should refer to the one found for normed spaces. Topology can be very abstract at first.
Looking at your two examples. In $\tau_1$, $(0,1)$ is open: elements in your topology are precisely those we declare open. Your confusion might come from the following situation: If you open a book, where $\mathbb{R}$ is considered, then one would typically say $(0,1)$ is open and not closed. However, one commonly look at $\mathbb{R}$ in the standard (i.e. metric) topology unless one mentions otherwise.
For $\tau_2$: In here $[0,1]$ is again open. This is not the standard topology by any means of course.
For the final bit: The discrete topology is somewhat one of the trivial ones. This is due to every possible singleton being open, so the topology will be the largest one possible. As such, being "open" in the discrete topology is not a special feature. Therefore, it is very unnatural to consider unless you work with intuitively discrete spaces such as $\mathbb{N}$.
Best Answer
In an abstract topological space, "open set" has no definition!
You simply decide (as part of making your topological space) which sets you want to call open -- those are the sets you put into $\mathcal T$. Whatever you decide to call open will be called open, as long as your decision meets the condition "$\emptyset, X\in\mathcal T$ and $\mathcal T$ is closed under arbitrary unions and finite intersections".
A metric space becomes a topological space by deciding that the "open sets" in this particular topological space are going to be exactly the ones that are open according to the metric-space definition.