Are open sets required to be in Topology T over set X? Assume that we choose two sets A and B to be open sets, contained in topology (X,T). A and B and the constructed sets that fulfill the requirements to make A and B open (unions and intersects of A, B, empty, and O) are represented by (X,O). Do all the constructed sets in (X,O) (sets other than A and B within (X,O)) have to be contained in the topology (X,T)?
 A: It is somewhat difficult for me to tell exactly what it is you are asking, but I think it's the following:

Suppose we have a topological space $(X,\tau)$, and we take $A$ and $B$ sets in $\tau$ (that is, open in the topology $\tau$). We then let $\mathcal{O}$ be the smallest collection of subsets of $X$ that includes $A$ and includes $B$, and is a topology on $X$ (so it is closed under arbitrary unions, finite intersections, contains $\emptyset$, contains $X$, contains $A$, and contains $B$). Is every set in $\mathcal{O}$ also in $\tau$?

The answer to this is yes. One nice way to see this is to consider the "topology generated by a collection of subsets of $X$". This is the $\mathcal{O}$ above; the difference is that I am describing it "from the top down", whereas you are attempting to describe it "from the bottom up". This is a very common situation in mathematics. So let's walk through the details.
Let $X$ be a set. Recall that a topology on $X$ is a collection $\tau$ of subsets of $X$ that satisfies the following three conditions:


*

*$\emptyset$ and $X$ are elements of $\tau$. ($\tau$ contains the empty set and the whole set).

*If $\{U_i\}_{i\in I}$ is a family of elements of $\tau$ (with $I$ arbitrary), then $\cup_{i\in I}U_i$ is also an element of $\tau$ ($\tau$ is closed under arbitrary unions).

*If $U_1$ and $U_2$ are in $\tau$, then $U_1\cap U_2$ is also in $\tau$ ($\tau$ is closed under finite intersections; the binary case yields the finite case by induction). 


The sets in $\tau$ are the "open sets in the topology".
Now suppose that we have two topologies on $X$, $\tau$ and $\sigma$. We have the following:
Lemma 1. If $\tau$ and $\sigma$ are topologies on $X$, then so is $\tau\cap\sigma$. That is, the collection of subsets of $X$ that are open in $\tau$ and also open in $\sigma$ form a topology.
Proof. We show that $\tau\cap\sigma$ satisfies the conditions to be a topology. First, since $\emptyset$ is in both $\tau$ and $\sigma$, it is in $\tau\cap\sigma$; likewise, since $X$ is in both $\tau$ and $\sigma$, it is in $\tau\cap\sigma$.
Now suppose that $\{U_i\}_{i\in I}$ is a collection of elements of $\tau\cap\sigma$. Then $\{U_i\}_{i\in I}$ are all in $\tau$, so $\mathscr{U}=\cup_{i\in I}U_i$ is in $\tau$. Since $\{U_i\}_{i\in I}$ are all in $\sigma$, then $\mathscr{U}=\cup_{i\in I}U_i$ is in $\sigma$ (since each of $\tau$ and $\sigma$ are closed under arbitrary unions). So $\mathscr{U}$ is in both $\tau$ and $\sigma$, hence in $\tau\cap\sigma$.
Finally, suppose that $U_1$ and $U_2$ are in both $\tau$ and $\sigma$ (that is, in $\tau\cap \sigma$. Since $\tau$ is closed under finite intersections, $U_1\cap U_2\in \tau$; since $\sigma$ is closed under finite intersections, then $U_1\cap U_2\in \sigma$. So $U_1\cap U_2\in \tau\cap \sigma$, showing that $\tau\cap\sigma$ is closed under finite intersections.
So, in summary, if $\tau$ and $\sigma$ are topologies on $X$, then so is $\tau\cap\sigma$. QED
In fact, there's nothing special about having two topologies. Try to prove the following:
Lemma 2. Let $X$ be a set, and let $\{\tau_j\}_{j\in J}$ be an arbitrary family of topologies on $X$. Then $\tau=\cap_{j\in J}\tau_j$ is a topology on $X$.
The proof is very similar to the one above.
One final observation: if $S$ is a collection of subsets of $X$ (topology or not), then there is at least one topology on $X$ that contains every set in $S$: the discrete topology, which is the topology in which every subset of $X$ is empty. So we have:
Theorem 1. Let $S$ be a collection of subsets of $X$. Let $\{\tau_j\}_{j\in J}$ be the family of all topologies on $X$ that contain $S$, that is, with $S\subseteq \tau_j$  (note that the family is non-empty). Then
$$\mathcal{O} = \bigcap \tau$$
is a topology on $X$ that contains $S$ (so $S\subseteq \mathcal{O}$). Moreover, $\mathcal{O}$ is the smallest topology on $X$ that contains $X$, that is, if $\sigma$ is any topology on $X$ such that $S\subseteq \sigma$, then $\mathcal{O}\subseteq \sigma$. 
Proof. Since $\mathcal{O}$ is an intersection of topologies (and there is at least one topology in the intersection, so it makes sense to take the intersection) then $\mathcal{O}$ is a topology on $X$ by Lemma 2 above.
Since $S\subseteq \tau_j$ for every $j$, then $S\subseteq \cap \tau_j=\mathcal{O}$, so $\mathcal{O}$ contains $S$. a
Finally, suppose that $\sigma$ is a topology on $X$ that contains $S$. Then, since $\{\tau_j\}_{j\in J}$ is the collection of all topologies on $X$ that contain $S$, then $\sigma=\tau_{j_0}$ for some $j_0$. So
$$\sigma = \tau_{j_0} \supseteq \cap_{j\in J}\tau_j = \mathcal{O},$$
hence $\mathcal{O}\subseteq \sigma$, as claimed. Thus, $\mathcal{O}$ is the smallest topology that contains $S$. QED
This theorem describes the smallest topology "top-down": by taking all possible topologies and "cutting them down" until we get the smallest possible topology that includes $S$. This is called the "topology generated by $S$".
For your situation, we can take $S=\{A,B\}$. Then $\mathcal{O}$ is the topology generated by $A$ and $B$, and in particular, any topology that has $A$ and $B$ in it will contain $\mathcal{O}$; since $\tau$, the topology you started with, contains $S$, then by the theorem we conclude that $\mathcal{O}\subseteq \tau$; that is, every set that is open in $(X,\mathcal{O})$ is also open in $(X,\tau)$, as you ask.
However, you might wonder if the topology $\mathcal{O}$ that I described is really the topology you described. I described the topology as the "intersection of all topologies on $X$ that contain $A$ and $B$". But the topology you described was the topology "with all sets that can be 'constructed' from $A$ and $B$" etc. That is, you want to consider the collection of all sets that can be formed by taking arbitrary unions and intersections, starting with $X$, $\emptyset$, $A$, and $B$, until you get a topology. This is a "bottoms up" description of the topology. 
In fact, there is a precise description of the sets we can "construct" from this set:
Theorem 2. Let $S$ be a collection of subsets of $X$ that includes $X$. Then the smallest topology on $X$ that contains $S$ is the collection of all subsets of $X$ that can be written as an arbitrary union of finite intersections of elements of $S$. 
Proof. Let us call the collection described $\mathcal{S}$. If $\sigma$ is any topology on $X$ that contains $S$, then it must include any finite intersection of elements of $S$, and so must contain any arbitrary union of finite intersections of elements of $S$. That is, $\mathcal{S}$ is contained in every topology that contains $S$. Therefore, $\mathcal{S}$ is contained in the smallest topology on $X$ that contains $S$.
We want to prove the converse: that the smallest topology on $X$ that contains $S$ is in fact contained in $\mathcal{S}$. To do this, it is enough to show that $\mathcal{S}$ is itself a topology on $X$ that contains $S$, because the smallest topology on $X$ that contains $S$ is contained in any topology that contains $S$.
Note that each element of $S$ can be written as itself, which is a union (of one set) of a finite intersection (of one set) of elements of $S$. So $S\subseteq \mathcal{S}$. Also, the empty set is the result of taking a union with no elements, so $\emptyset\in\mathcal{S}$. And since $X\in S$ by hypothesis, then $X\in\mathcal{S}$.
Let $\mathcal{A}_{i\in I}$ be a collection of unions of finite intersections of elements of $S$. Then $\cup_{i\in I} \mathcal{A}_i$ is a union of unions of finite intersections of elements of $S$; since a union of unions is itself a union, $\cup_{i\in I} \mathcal{A}_i$ is a union of finite intersections of elements of $S$, hence an element of $\mathcal{S}$. So $\mathcal{S}$ is closed under arbitrary unions.
To prove that $\mathcal{S}$ is closed under finite intersections, suppose that
$\mathcal{U}_1$ and $\mathcal{U}_2$ are each arbitrary unions of finite intersections of elements of $S$; that is,
$$\begin{align*}
\mathcal{U}_1 &= \cup_{i\in I}(U_{i1}\cap\cdots\cap U_{in_i})\\
\mathcal{U}_2 &= \cup_{j\in J}(U_{j1}\cap\cdots\cap U_{jn_j}),
\end{align*}$$
with $U_{rs}\in S$ for each pair $r$, $s$.
Then, since unions distribute over intersections, we have:
$$\begin{align*}
\mathcal{U}_1\cap\mathcal{U}_2 &= \left(\cup_{i\in I}(U_{i1}\cap\cdots\cap U_{in_i}\right) \cap \left(\cup_{j\in J}(U_{j1}\cap\cdots\cap U_{jn_j}\right)\\
&= \cup_{i\in I}\left( (U_{i1}\cap\cdots\cap U_{in_i})\cap\cup_{j\in J}(U_{j1}\cap\cdots\cap U_{jn_j})\right)\\
&= \cup_{i\in I}\cup_{j\in J}\left(U_{i1}\cap\cdots\cap U_{in_i}\cap U_{j1}\cap\cdots \cap U_{jn_j}\right).
\end{align*}$$
This is an arbitrary union of finite intersections of elements of $S$, so it is in $\mathcal{S}$. That is, $\mathcal{S}$ is closed under finite intersections.
So $\mathcal{S}$ is a topology; it contains $S$, so it contains the smallest topology that contains $S$. But it is also contained in the smallest topology that contains $S$, so $\mathcal{S}$ is the smallest topology that contains $S$. QED
We say that $S$ is a subbase of the topology $\mathcal{S}$. In general, given a topology $\tau$ and a set $S$, if every element of $\tau$ is an arbitrary union of finite intersections of elements of $S$ (together with $X$ itself), then $S$ is said to be a subbase for $\tau$.
So the topologies described in Theorem 1 and Theorem 2 are really the same topology. Theorem 1 describes the topology by a "universal property": a topology satisfying certain property (includes every element of $S$) and in a particular relation with every other topology that has that property (in this case, contained in any other topology that contains $S$). Theorem 2 describes the topology "from the bottom up", by describing how to obtain the elements of the topology in terms of the elements of $S$.
It is usually best to have both descriptions for such a construction; the universal property definition usually makes proving things about the topology, relative to other topologies, easy (for example, your question is almost immediate with Theorem 1, but pretty hard to prove directly with Theorem 2); the explicit description actually takes some of the "mystery" out of the topology and lets you get your hands very explicitly on the elements of that topology.
So, in summary, after a lot, and assuming I understood your question, "definitely yes: every open set in the topology generated by $A$ and $B$ is an open set in the topology $\tau$, because $A$ and $B$ are open sets in $\tau$."
A: As an example demonstrating that two topologies need not be comparable, let $X = \{a, b\}$ (a set with just two elements). Define on $X$ the topologies
$$
\sigma = \{\emptyset, \{a\}, X\}
$$
and
$$
\tau = \{\emptyset, \{b\}, X\}.
$$
(the sets listed are taken to be the open sets of the topology). Both $(X, \sigma)$ and $(X, \tau)$ are topological spaces (you can check that they satisfy the axioms), but neither one is contained in the other.
A: If (X,O) is a topology which includes open sets A and B and constructed sets, then all (X,O) is contained in (X,T). Therefore (X,O) is a subset of (X,T).
Prove (X,O) is contained in (X,T):
If two sets A and B are considered open and contained in (X,T), all their constructed sets, which are required for A and B to be open, would also be contained in (X,T). Since (X,T) contains A and B, all unions and intersects of A, B, empty, and X are in (X,T). This is because the requirements for a set to be open are identical to the requirements for a set to be in a topology.
