The question you linked to essentially asks, "Is every sigma algebra a topology?" Since the answer is no, the OP was given a counterexample.
I'm not sure that this really is a "basic question". It is more of a question coming from one noticing that the axioms for a sigma algebra and the axioms for a topology share a lot of similarities.
However, topologies and sigma algebras are built to accomplish very different tasks. Topologies are defined in order to capture some abstract version of "closeness" so that one can study continuity. This works best if we allow arbitrary unions of open sets to be open.
On the other hand, sigma algebras are built to provide a collection of sets to which we can attach a number (a measure). But there is a delicate balance, we want to include a rich enough collection so we can measure as many things as possible. But on the other hand, we need to be careful about infinite sums. In the end, uncountable infinite sums cause problems: See this question. We really want to stick to countably indexed sums. Thus the limited unions.
So in the end sigma algebras and topologies are different because they are tools for dealing with different sets of problems.
As for your second question, it's not entirely clear what you're asking. I'll make a stab at answering what I think you're asking...
The singleton sets $\{ x_0 \}$ are Borel subsets on $I=(0,1)$ (where $0<x_0<1$). If the Borel sets formed a topology, then arbitrary unions of these singleton sets would have to belong to the collection of Borel sets (since topologies are closed under arbitrary unions). This would imply that all subsets of $I=(0,1)$ are Borel sets (since every set is a union of singleton sets)! However, there are subsets of $I=(0,1)$ which fail to be Borel sets. Thus the Borel sets of $I=(0,1)$ do not form a topology (even though they form a $\sigma$-algebra).
