Here the solution of Problem 5(b) of Chapter 2 in Munkres's book Topology, the solution is given as follows (I will be copying exactly the solution given in that book just for the sake of completeness of the question),
Problem.
Let $X$ and $X'$ denote a single set in the topologies $\mathscr{T}$ and $\mathscr{T}'$ , respectively; let $Y$ and $Y'$ denote a single set in the topologies $\mathscr{U}$ and $\mathscr{U}'$, respectively. Assume these sets are nonempty.
(a) Show that if $\mathscr{T}'\supseteq \mathscr{T}$ and $\mathscr{U}'\supseteq \mathscr{U}$ , then the product topology on $X'×Y'$ is finer than the product topology on $X×Y$.
(b) Does the converse of (a) hold? Justify your answer.
Solution. (I won't reproduce the solution of 5(a) here because I think it's all right.)
(b) Yes (assuming the sets are nonempty). If $U$ is open in $X$, $x∈X$, $V$ is open in $Y$, $y∈Y$ , then $U×V$ is open in $X×Y$ and, therefore, open in $X'×Y'$ . Therefore, there exists a basis set $A×B$ in $X'×Y'$ such that it is a subset of $U×V$ and it contains $x×y$. Therefore, there are open sets $A∈\mathscr{T}'$ and $B∈\mathscr{U}'$ such that $x∈A⊂U$ and $y∈B⊂V$ . So, $U$ is open in $X'$ and $V$ is open in $Y'$ .
Now my question is that where in the proof we show that $\mathscr{T}\subseteq \mathscr{T}'$ and $\mathscr{U}\subseteq \mathscr{U}'$? What we have shown is simply the following,
For all sets $U$ open in $X$ and $V$ open in $Y$, $U$ is open in $X'$ and $V$ open in $Y'$.
But that doesn't prove that $\mathscr{T}\subseteq \mathscr{T}'$ and $\mathscr{U}\subseteq \mathscr{U}'$. What about all those sets $U\in\mathscr{T}$ such that $X\subset U$?
Am I missing something?
The question was clarified in the comments. Here is the clarification -
$X$ is your whole space. The only open set that contains $X$ is itself. So you don't really have any other sets $U\in\mathscr T$ such that $X\subset U$
The question says $X$ and $X′$ are a single set in the topologies $\mathscr T$ and $\mathscr T′$. What they are trying to say is that $X$ and $X′$ are actually the same set. When you give this set the topology $\mathscr T$ you call it $X$ and when you give it the topology $\mathscr T′$ you call it $X′$. They are are not asking you to take one element from the collections $\mathscr T$ and $\mathscr T′$ respectively.