The topology generated by a basis is defined as $\tau=\{U\subset X|\forall x\in U, \exists \beta\subset B, x\in \beta\}$. It is also true that all sets in the topology are unions of basis sets. I have a few questions on this:
1) Consider the Basis $B=\{(a,b)| a<b \in \Bbb Q\}$, I know that this generates the standard topology $\tau_{st}=\{ \cup(a,b)|a<b\in \Bbb R\}$. But consider two disjoint sets in $B$ say $(1,2),(7,8)$ Then their union is not an open interval so is not of the form $(a,b)$, so when we say that every open set is a union of basis sets are we saying that although all sets in $\tau$ can be formed by unions of basis sets not all unions of basis sets will form open sets in $\tau$?
2)As we see above the union symbol has been dispensed with , which makes sense in that we can union open intervals to get another open interval. Can we always (or at least usually dispense with this union symbol when using it in proofs, ( like if we wanted to find the closure of a set and we knew its basis would we usually have to keep referring or not to the elements of $\tau$ in terms of unions of basis sets) Can we usually do this ? for instance here Did I show that this is a basis for a topology on $X=\Bbb N \cup \{-1/n | n \in \Bbb N \}$ correctly? , the basis elements are the unions of rays and intervals, so unions of these intervals would again be rays and intervals and we could once again dispense with the union symbol ? If I am correct is saying we can usually dispense with the union symbol then is there an example of a time when we would not be able to dispense with it ?
3) additionally (although this is not in relation to the topology generated by the basis but is rather just a question about the basis itself ) If we are given a Basis that is formed through unions e.g. $B=\{(a,\infty) \cup (-1,b)\cup (c,d) |a,b,c,d \in \Bbb \}$ and $X=\Bbb N \cup \{-1/n|n\in \Bbb N\}$ then say we are trying to see if we can find a basis set for every $x\in X$ (i.e. trying to see if $X$ is covered by the basis , then say $x\in \Bbb N$ $x\neq 1 $ do we write $x \in (a,\infty)\in B$ or do we have to write $x \in (a,\infty)\cup (-1,b)\cup (c,d) \in B$, in other words are the elements always the unions or could we just take the other two sets to be empty and say that $(a,infty)$ by itself is also an element ?
I'm not really sure what $\cup(a,b)$ is supposed to represent. $\cup$ is a binary operation (where $x\cup y=\{\alpha:\alpha\in x\text{ or }\alpha\in y\}$) or else you're taking the union of a family (where $\bigcup X=\{\alpha:\text{for some }x\in X\text{, }\alpha\in x\}$). Regarding the interval $(a,b)$ as a family of sets would be... odd, and it require you to get into how exactly the reals are constructed.
You seem to be confused about what an open vs. base open set is. A basis is a collection of open sets that will give rise to the rest of the open sets. But just because something is an open set, that doesn't mean that it is a base open set. It just means that opens sets can be decomposed into base open sets, and in particular each open set is the union of base open sets. For example, with base open sets being open intervals of $\mathbb{R}$, $(0,1)\cup(1,2)$ is an open set, but is not an open interval.
$(1,2)\cup(7,8)$ is open, because it is the union of two of the base open sets: namely it is the union of $(1,2)$ and $(7,8)$. $(1,2)\cup(7,8)$ is not an element of the basis, but it is open.
I'm not really sure what you're asking here. If you have another way of describing things without unions, then you don't need to use unions. So metric spaces, for example, might define things in terms of their metric rather than unions of base open sets. Sometimes the generated topology will use unions in its definition just because it makes the notion of an open set a little more concrete: we have these base open sets as given, and all the other open sets are just unions of these. The end result should always be equivalent regardless of what symbols you use. It would just be two ways of representing the same thing.
I think you may have a typo in the definition of the generated topology: you need $\beta\in B$ not $\beta\subseteq B$. And I think you may also have a misunderstanding about it. Being covered by a basis set doesn't tell you whether you're open or not. If you're trying to show that $X$ is open (and I'm not really sure what you're asking), you would need to show that every $x\in X$ has a base open set $N$ with $x\in N\subseteq X$. In your example, your basis elements are always unions, but if $a<-1$, you can have $(a,\infty)\in B$ since taking $c=-1<d=1$ yields that $(-1,b)\subseteq (c,d)\subseteq(a,\infty)$ and hence $(a,\infty)\cup(-1,b)\cup(c,d)=(a,\infty)$.