Why the collection of all closed sets is not a topology? Here, what I mean closed and open sets are those in real analysis sense. $[a,b]$ is a closed interval, and $(a,b)$ is an open interval.
My reasoning:
By the axioms for a topology, any infinite or finite union of members in a topology needs to be in the topology. On another hand, only finite intersection of members in the topology needs to be in the topology.
In one dimension, I may construct an infinite union of closed intervals that approaches an non-closed interval.  Such as $\cup_{x\in N} [0, (2x-1)/x]$ that approaches $[0,2)?$
For open interval to do the same, we need infinite intersections instead of infinite unions, but infinite intersections are not required to be in the topology. Is this why the closed intervals do not construct a topology but  the open intervals do? 
 A: 
Is this why the closed intervals do not construct a topology but the open intervals do?

Not really. Neither closed nor open intervals form a topology. Is $[0,1]\cup[3,4]$ an interval? Or $(0,1)\cup(3,4)$?
So I assume that what you are asking is "is this why closed subsets do not form a topology unlike open subsets"? I assume that by "analytic sense" you mean "induced by the Euclidean metric". Otherwise a question "why open subsets form a topology?" is a bit weird - by definition.
Then your reasoning is almost correct regarding closed subsets. First of all a closed/open set need not be a closed/open interval. So you got the first part right:
$$\bigcup [0,(2x-1)/x]=[0,2)$$
however here's where you are wrong: you say $[0,2)$ is not an open interval but this doesn't mean that it is not open. For example $(0,1)\cup(3,4)$ is an open subset that is not an open interval.
But $[0,2)$ indeed is not closed (in Euclidean sense) hence it cannot be open in our new topology. It is not closed because the very same sequence $(2x-1)/x$ is convergent outside of it.
A: The open intervals form a base for the (usual) topology on $\mathbb{R}$, because every open set is a union of open intervals.
A family of subsets of $X$ is the base for a topology on $X$ if and only if


*

*the family is a cover for $X$ (the union of the family is $X$);

*the intersection of two members of the family is a union of members of the family.


If a family satisfies these property, then the set of arbitrary unions of members of the family is a topology on $X$.
We clearly see that the open intervals satisfy both conditions (the second almost trivially, because the intersection of two open intervals is either empty or an interval).
Do the closed intervals form a base for a topology? Certainly so, if we admit that singletons are closed interval; but in this case the induced topology is the discrete topology.
To the contrary, the topology having as a base the set of half-open intervals $[a,b)$ is quite interesting: it is called the Sorgenfrey line.
