The lub of the usual and co-countable topology on $\mathbb R$ This exercise comes from Steven A. Gaal's Point Set Topology(1964,Academic Press), Page 37, exercise 4.  It is


  
*Let $\mathscr T$ be the least upper bound of the usual topology and of the topology of countable complements on the set of reals.  Show that $O$ is open relative to $\mathscr T$ if and only if $O=Q-A$ where $Q$ is open in the usual sense and $A$ is countable.
  

There are some concepts need to be clarify:


*

*The open sets of the topology of countable complements are $\varnothing$ and those sets $A$ whose complement $cA$ is countable.

*The least upper bound of a family of topologies on $X$ is the least fine topology on $X$ which is finer than any topology in this family.
Considering all of them above, the exercise is equivalent to:

Let $\mathscr T_1$ be the standard topology on the real line, and let $\mathscr T_2$ be the cocountable topology on the real line.  Consider the topology $\mathscr T$, which is generated by the subbase $\mathscr T_1 \cup \mathscr T_2$.  Prove that $U\in \mathscr T$ if and only if $U=V\setminus A$ where $V\in \mathscr T_1$ and $A$ is countable.

This exercise frustrates me and I am very eager to know the solution.  I have never seen this exercise before, and after browsing and searching on various books and resources that I can have, I finally come to ask for the answer here.  Any help will be appreciated, thank you! 
 A: Since the OP states 'Any help will be appreciated...', I offer the following without apologies.
Consider the deleted integer topology on $X = \{x \in \mathbb R \, | \, x \gt 0 \text{ and } x \notin \mathbb N \}$. The set $X$ can also be endowed with the cocountable topology.
Exercise: Describe the open sets in the topology $\mathscr T$ that is the least upper bound of the deleted integer topology and the cocountable topology on $X$.
Hint: The deleted integer topology has a countable basis.
Instead of solving this exercise, we content ourselves here by defining a topology on $X$ by specifying a base of 'slightly altered' open intervals:
$\tag 1 \mathscr B = \{ (n - 1, n) - C_\delta \, | \, n \ge 1 \text{ and } C_\delta \subset (n-1,n) \text{ is any countable set} \}$
Since the intervals $(n-1, n)$ partition $X$ and the $C_\delta$ can be empty, $\mathscr B$ is a cover for $X$. If we take two sets in $\mathscr B$ the intersection is empty if they are not formed from the same $n \ge 1$. But if we intersect two with the same $n$,
$\tag 2 [(n - 1, n) - C_{\delta}] \; \bigcap \; [(n - 1, n) - C_{\delta^*}] = (n - 1, n) - (C_{\delta} \cup C_{\delta^*}) $
Since the union of two countable sets is countable, $\mathscr B$ is closed under the intersection operator.
We have shown that that $\mathscr B$ forms a base for a topology on $X$.
More Hints/Facts Regarding the Exercise:
Note first that an arbitrary union of sets in $\mathscr B$ can be written as a countable union by collecting the base sets under a given $(n-1,n)$ block (each $(n-1,n)$ collection belonging to $\mathscr B$). Also, any countable set $C \subset X$ can be expressed as a union of countable sets, $C \cap (n-1,n)$. Finally, for any collection of countable sets $C_n \subset (n-1,n)$, the countable union, contained in $X$, is countable. 

The argument given here, in this simpler setting, can be reworked to characterize the open sets in $\mathbb R$ with the OP's 'lub topology'.
