I like to think of topological spaces as defining "semidecidable properties". Let me explain.
Imagine I have an object that I think weighs about one kilogram. Suppose that, as a matter of fact, the object weighs less than one kilogram. Then I can, using a sufficiently accurate scale, determine that the object weighs less than one kilogram. Even if the object weighs, say, 0.9999996 kilograms, all I need to do is find a scale that's accurate to within, say, 0.0000002 kilograms, and that scale will be able to tell me that the object weighs less than one kilogram.
This means that "weighing less than one kilogram" is a semidecidable property: if an object has the property, then I can determine that it has the property.
Suppose, on the other hand, that the object actually weighs exactly one kilogram. There's no way I can measure the object and determine that it weighs exactly one kilogram, because no matter how precisely I measure it, it's still possible that there's some amount of error which I haven't discovered yet. So "weighing exactly one kilogram" is not a semidecidable property.
What does this have to do with topological spaces? Well, an open set in a topological space corresponds to a semidecidable property of that space. This is why in the topological space of real numbers, the set $\{x : x \in \mathbb{R}, x < 1\}$ is an open set, but the set $\{x : x \in \mathbb{R}, x = 1\}$ is not.
So, consider the "topological space" $X = \{a, b, c\}$ with open sets $\emptyset$, $\{a, b\}$, $\{b, c\}$, and $\{a, b, c\}$. In this "topological space", you are asserting that
- (since $\{a, b\}$ is open) if you have a point which is either $a$ or $b$, then it is possible to measure it and determine that it is either $a$ or $b$ (though it is not necessarily possible to determine which one it is);
- (since $\{b, c\}$ is open) if you have a point which is either $b$ or $c$, then it is possible to determine that it is either $b$ or $c$; but
- (since $\{b\}$ is not open) if you have the point $b$, it is not possible to determine that it is $b$.
However, these assertions contradict each other. Suppose that you have the point $b$. Because of the first bullet point, there is some measurement you can make which will tell you that the point is either $a$ or $b$. And because of the second bullet point, there is another measurement you can make which will tell you that the point is either $b$ or $c$. If you simply make both of these two measurements, then you will have successfully determined that the point is (either $a$ or $b$, and either $b$ or $c$)—in other words, that the point is $b$. But the third bullet point asserts that this is impossible!
For more explanation of this idea, see these two answers:
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
About the last question: the choice to start with $\tau$ as the collection of open subsets is arbitrary. You could take the collection of closed subsets and you would still get the same information. This is because closed and open subsets are put in bijection by taking the complement. More precisely if you have $\tau$ a family of subsets satisfying the usual axioms of basis then $\{A^c : A \in \tau \}$ would give you a family of subsets satisfying a set of axioms complementary to the former (i.e. we ask now closure under arbitrary intersections and only finite unions).
Regarding the first and more interesting question: I think that the measure of how much two subsets are close is captured by their topological closure. In a metric space $(X,d)$ the usual definition of distance between two subsets $A,B$ is given by the formula $d(A,B)=\inf_{a \in A, b \in B}d(a,b)$. This in fact is the same as the distance between their closures $\overline{A}, \overline{B}$. Take as explicit example $\mathbb{R}$ with the usual distance: you have that a real number $r$ is intuitively close to a subset iff it has distance zero from a set. Let's take an interval $I=(a,b)$ with $b \leq r$: then the distance between $r$ and $I$ is given by $|b-r|$. Observe that the distance does not differentiate between the cases that $I$ is open or closed, in fact $d(I,r)=d(\overline{I},r)$.
The idea is that on a general topology $(X, \tau)$ you can distinguish two elements $x,y \in X$ if you can provide open subsets containing only one of the two: in the case of a metric space you have open balls giving you disjoint open neighborhoods separating the two, but in general spaces the situation is much more complicated and there are a lot of different separation axioms stating how much reasonably you can ask two points to be distinct.
Regarding the two examples of different topologies on the same set $X$: in the first case you can easily see that $\{c \}$ is closed while $\overline{\{a \}}=\{ a,c\}$ and similarly $\overline{\{b \}}=\{ b,c\}$. You have a situation where $c$ cannot be separated from the other elements since you do not have enough open subsets. For the second topology you have now $\overline{\{a \}}=X$: you could interpret this by saying that since you removed open subsets from your stating topology $\tau_1$ you have less information so you cannot distinguish $a$ from the rest of the set $X$ as well as before.