Theorem 16.3 of Munkres Topology 
If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A\times B$ is the same as the topology $A\times B$ inherits as a subspace of $X\times Y$.

Since I didn’t known the basis for subspace topology on $A\times B$, I tryed to show two topology are equal, i.e.$\mathcal{T}_{p}$(product topology on $A\times B$. Note talking about product topology makes sense, because set $A$ and $B$ are equipped with $\mathcal{T}_{A}$ and $\mathcal{T}_{B}$ topology, respectively)$=\mathcal{T}_{A\times B}$(subspace topology on $A\times B$)
My attempt: first I will show notation. $\mathcal{B}_{p}=\{ U\times V|U\in \mathcal{T}_{A}, V \in \mathcal{T}_{B}\}$. $\mathcal{T}_{p}$ is the topology generated by basis $\mathcal{B}_{p}$. $\mathscr{B}_{p}$ is basis for product topology on $X\times Y$. $\mathcal{T}_{A\times B}= \{ (A\times B)\cap Z| Z \in \mathcal{T}_{X\times Y}\}$, where $\mathcal{T}_{X\times Y}$ is the product topology on $X\times Y$. I know there is lots of notation to keep track of. If $R\in \mathcal{T}_{p}$, then $R=\bigcup_{i\in I}(U_i \times V_i)$, where $U_i\times V_i \in \mathcal{B}_{p}, \forall i \in I$ & $I$ is an indexing set. Since $U_i \times V_i \in \mathcal{B}_{p}$, we have $U_i\in \mathcal{T}_{A}$ and $V_i\in \mathcal{T}_{B}$. So $U_i=A\cap E_i$ & $V_i=B\cap F_i$, $E_i\in \mathcal{T}_{X} $ and $F_i \in \mathcal{T}_{Y}$. So $U_i \times V_i =(A\cap E_i)\times (B\cap F_i)= (A\times B)\cap (E_i \times F_i), \forall i \in I$. Since $E_i \times F_i \in \mathscr{B}_{p}$, $E_i \times F_i \in \mathscr{B}_{p} \subset \mathcal{T}_{X\times Y}, \forall i$. So $E_i \times F_i \in \mathcal{T}_{X\times Y}, \forall i$. So $\bigcup_{i\in I} (E_i \times F_i) \in \mathcal{T}_{X\times Y}$. Thus, $R=\bigcup_{i\in I}(U_i \times V_i)= (A\times B) \cap ( \bigcup_{i\in I} E_i \times F_i)\in \mathcal{T}_{A\times B}$. Hence $\mathcal{T}_{p} \subseteq \mathcal{T}_{A\times B}$.
Conversely, suppose $R\in \mathcal{T}_{A\times B}$. Then $R=(A\times B) \cap Z$; $Z \in \mathcal{T}_{X\times Y}$. $\forall (m,n)\in Z, \exists W_{(m,n)}\in \mathscr{B}_p$ such that $(m,n)\in W_{(m,n)}\subseteq Z$. So $Z = \bigcup_{(m,n)\in Z} W_{(m,n)}$. $R=(A\times B)\cap (\bigcup_{(m,n)\in Z} W_{(m,n)})= \bigcup_{(m,n)\in Z}[(A\times B)\cap (W_{(m,n)})]$. Let $W_{(m,n)}=U_{(m,n)}\times V_{(m,n)} \in \mathscr{B}_{p}$. So $R= \bigcup_{(m,n)\in Z}[(A\cap U_{(m,n)})\times (B\cap V_{(m,n)})]$. Since $U_{(m,n)}\in \mathcal{T}_{X}$ and $V_{(m,n)}\in \mathcal{T}_{Y}$, we have $A\cap U_{(m,n)}\in \mathcal{T}_{A}$ and $B\cap V_{(m,n)}\in \mathcal{T}_{B}$. Thus $(A\cap U_{(m,n)})\times (B\cap V_{(m,n)})\in \mathcal{B}_{p}, \forall (m,n)\in Z$. Hence $R$ is the arbitrary union of elements of $\mathcal{B}_{p}$. So $R\in \mathcal{T}_{p}$. Thus $\mathcal{T}_{A\times B} \subset \mathcal{T}_{p}$. We have our desired equality $\mathcal{T}_{A\times B} = \mathcal{T}_{p}$. Is this proof correct?
Question: What is the basis for subspace topology on $A\times B$? Is it the set itself? I don’t understand Munkres proof of, basis of product topology on $A\times B$ $\subseteq$ basis of subspace topology on $A\times B$.

Munkres’ Proof: $\mathcal{B}_{p} =\{ U\times V|U\in \mathcal{T}_{A}, V \in \mathcal{T}_{B}\}$(basis of product topology on$A\times B$) is equal to $\mathcal{B}_{s}=\{(A\times B) \cap O|O\in \mathcal{B}\}$(basis of subspace topology on $A\times B$), where $\mathcal{B}=\{ C\times D|C\in \mathcal{T}_{X}, D\in \mathcal{T}_{Y}\}$. We need to show that $\mathcal{B}_{p}= \mathcal{B}_{s}$. If $U’\in \mathcal{B}_{p}$, then $U’=U\times V$. So $U=A\cap M$ & $V=B\cap N$, where $M\in \mathcal{T}_{X}$ and $N\in \mathcal{T}_{Y}$. $U’=U\times V=(A\cap M)\times (B\cap N)=(A\times B)\cap (M\times N)\in \mathcal{B}_{s}$. Thus $\mathcal{B}_{p} \subseteq \mathcal{B}_{s}$.
Conversely, if $U’\in \mathcal{B}_{s}$, then $U’=(A\times B)\cap O$; $O\in \mathcal{B}$. So $U’=(A\times B)\cap (P\times Q)$, where $P\in \mathcal{T}_{X}$ & $Q\in \mathcal{T}_{Y}$. $U’=(A\cap P)\times (B\cap Q) \in \mathcal{B}_{p}$, since $(A\cap P)\in \mathcal{T}_{A}$ and $(B\cap Q)\in \mathcal{T}_{B}$. Hence  $\mathcal{B}_{p} \supseteq \mathcal{B}_{s}$. Thus $\mathcal{B}_{p}= \mathcal{B}_{s}$. So topology generated by each basis are equal. Our desired result.
 A: I personally think the following is a more readable version of such a write-up:
$A$ and $B$ have the subspace topologies $\mathcal{T}_A$ resp. $\mathcal{T}_B$ inherited from $X$ resp. $Y$.
Suppose $O$ is open in $A \times B$ in the product topology $\mathcal{T}_A \times \mathcal{T}_B$. So $O= \bigcup_{i \in I} (U_i \times V_i)$, where $U_i$ open in $A$, $V_i$ open in $B$, for all $i \in I$.
By the definition of the subspace topology on $A$ resp. $B$, we write $U_i = \hat{U}_i \cap A$, $V_i = \hat{V}_i \cap B$ for open $\hat{U}_i$ in $X$ and open $\hat{V}_i$ in $Y$ for all $i$.
Then $\hat{O} = \bigcup_{i \in I} (\hat{U}_i \times \hat{V}_i)$ is open in $X \times Y$ in the product topology and $$\hat{O} \cap (A \times B) = (A \times B) \cap \bigcup_{i \in I} (\hat{U}_i \times \hat{V}_i) = \bigcup_{i \in I} \left((\hat{U}_i \cap A) \times (\hat{V}_i \cap B)\right)  =\bigcup_{i \in I}(U_i \times V_i) = O$$
So $\mathcal{T}_A \times \mathcal{T}_B \subseteq (\mathcal{T}_X \times \mathcal{T}_Y)_{A \times B}$.
On the other hand, if $O$ is open in $(\mathcal{T}_X \times \mathcal{T}_Y)_{A \times B}$, we write $O = \hat{O} \cap (A \times B)$ where $\hat{O}$ open in $\mathcal{T}_X \times \mathcal{T}_Y$, so $\hat{O} = \bigcup_{i \in I} (U_i \times V_i)$, with $U_i$ open in $X$, $V_i$ open in $Y$. Again we note that
$$O = \hat{O} \cap (A \times B) = \left(\bigcup_{i \in I} (U_i \times V_i)\right) \cap A \times B = \bigcup_{i \in I} ((U_i \cap A) \times (V_i \cap B))$$
which is a union of open sets of $\mathcal{T}_A \times \mathcal{T}_B$, so in that topology. This proves the other inclusion.
For a completely abstract and more general statement from which this also follows (which I learnt in my first course of general topology as the transitive law of initial topologies, see my answer here.) Not for the faint of heart...
A: Abstract Mindset
It is usually better to try to prove things (especially in  more modern subjects of mathematics) in a more abstract way. At some point the sets become too complicated to deal with. Especially if it is sets, within sets, within sets, ect. So one should try to a more categorical mindset.
Sometimes one has to "get dirty" and write everything out explicity. But often times it is completely unnecessary. As a result one can get the same result by appealing to something deeper. Here is an alternative approach that seems to work.

Sub-Space Topology
Let $X$ be a (topological) space and $A$ a subset of $X$. The sub-space topology of $A$ can be characterized by the following properties:

*

*The inclusion map $i:A\to X$ is continuous.


*If $A$ is given any topology, say we call it by $A'$ (to avoid confusion, $A'=A$ as a set, but with another topology), such that $i:A'\to X$ is continuous, then the identity map $A'\to A$ is continuous.

Product Topology
If $X$ and $Y$ are spaces then the product space $X\times Y$ can be characterized by the following two properties:

*

*The project maps $\pi_X: X\times Y\to X$ and $\pi_Y: X\times Y\to Y$ are continuous.


*If there is another space $P$ with a pair of continuous maps $f:P\to X$ and $g:P\to Y$, then there exists a unique continuous map $\phi:P\to (X\times Y)$ such that $\pi_X\circ \phi = f$ and $\pi_Y\circ \phi = g$.

The Proof
Suppose $A$ and $B$ are given the sub-space topology.
Since $A\to X$ and $B\to Y$ are continuous we know the "product map" $A\times B\to X\times Y$ is also continuous (I assume you understand this step, if not I can supply more details).
Now suppose that $(A\times B)'$ is the same set as $A\times B$ but with the subspace  topology inherited from $X\times Y$.
Since $(A\times B)\to (X\times Y)$ is continuous we know that the identity map $(A\times B)\to (A\times B)'$ is continuous from the first universal property.
Now we will argue, the other way, that the identity map $(A\times B)'\to (A\times B)$ is also continuous, thereby showing that the two topologies are the same.
To accomplish this we begin with the observation that $j:(A\times B)'\to (X\times Y)$ is continuous, since it has the sub-space topology. However, $\pi_X: (X\times Y)\to X$ and $\pi_Y:(X\times Y)\to Y$ are continuous. Therefore, $\pi_X\circ j:(A\times B)'\to X$ and $\pi_Y\circ j: (A\times B)'\to Y$ are also continuous.
Now because the images pf $\pi_X\circ j$ and $\pi_Y\circ j$ are just $A,B$, and $A,B$, have the sub-space topology themselves, it means $\pi_X\circ j:(A\times B)'\to A$ and $\pi_Y\circ j:(A\times B)'\to B$ are both continuous.
Therefore, from the second universal property, (for the product $A\times B$) there exists a unique continuous map $\phi:(A\times B)'\to (A\times B)$, such that $\pi_A \circ \phi = \pi_X\circ j$ and $\pi_B\circ \phi = \pi_Y \circ j$. Note, $\phi = \text{id}$ satisfies this condition, by uniqueness we have that $\phi = \text{id}$ and so $\text{id}:(A\times B)'\to (A\times B)$ is continuous.
