Consider $X=Y=\mathbb{R}$ and $A=B=(-\infty,0)\cup (0,\infty).$ We have that
$$(A\times B)^c=\{(x,y)\in\mathbb{R}^2|xy=0\}$$ is not finite.
You can of course define many topologies $\tau_Y$ on the set $Y$, but a minimal requirement is certainly that the inclusion function $i : (Y,\tau_Y) \to (X,\tau_X)$ becomes continuous. Let us call such a topology admissible (this is just an ad hoc notation).
Obviously there exist a finest admissible topology $\tau_Y^f$ (the discrete topology) and a coarsest admissible topology $\tau_Y^c$ (take the intersection of all admissible topologies). Moreover, the set $\tau_Y^s = \{U \cap Y \mid U \in \tau_X\}$ occurring in your question is easily seen to be a topology on $Y$; it is admissible because $i^{-1}(U) = U \cap Y \in \tau_Y^s$ for all $U \in \tau_X$.
We have $\tau_Y^c = \tau_Y^s$. The inclusion $\tau_Y^c \subset \tau_Y^s$ holds by definition, and since $\tau_Y^c$ is admissible, we have $U \cap Y = i^{-1}(U) \in \tau_Y^c$ for all $U \in \tau_X$, thus $\tau_Y^s \subset \tau_Y^c$.
This explains the special role of $\tau_Y^s$ among admissible topologies: It is the coarsest admissible topology on $Y$.
Let us now change perspective. We call a topology $\tau_Y$ on $Y$ subspace topology if it has the following universal property:
A function $f : (Z,\tau_Z) \to (Y,\tau_Y)$, where $(Z,\tau_Z)$ is any topological space, is continuous if and only $i \circ f : (Z,\tau_Z) \to (X,\tau_X)$
is continuous.
That is, the continuity of functions into $(Y,\tau_Y)$ can be detected by regarding them as functions into the ambient $(X,\tau_X)$. This is a very reasonable approach to introduce a topology on $Y$.
Each subspace topology $\tau_Y$ is admissible:
The identity $id: (Y,\tau_Y) \to (Y,\tau_Y)$ is continuous, thus $i = i \circ id$ is continuous.
There exists at most one subspace topology on $Y$:
If $\tau_Y, \tau'_Y$ are subspace topologies, then both $id : (Y,\tau_Y) \to (Y,\tau'_Y)$ and $id : (Y,\tau'_Y) \to (Y,\tau_Y)$ are continuous by 1. and the universal property. This means $\tau_Y = \tau'_Y$.
$\tau_Y^s$ is a subspace topology (and thus by 2. the unique subspace topology):
The "only if" part is trivial because $\tau_Y^s$ is admissible and for each admissible topology the continuity of $f$ implies the continuity of $i \circ f$. Let us prove the "if" part. Continuity of $i \circ f$ means that $(i \circ f)^{-1}(U) \in \tau_Z$ for all $U \in U_X$. But $(i \circ f)^{-1}(U) = f^{-1}(i^{-1}(U)) = f^{-1}(U \cap Y)$, thus $f^{-1}(V) \in \tau_Z$ for all $V \in \tau_Y^s$, i.e. $f$ is continuous.
Also have a look at an injective map is embedding iff its left composition with any continuous map is continuous.
If you know some category theory, Attempt to define the notion of subobjects will give you a broader understanding.
Best Answer
You are considering subspace topologies on what subsets? $X$ and $Y$ are not subsets of their cartesian product; you could embed them via some injection, for which on the other hand there is no canonical choice (and would not even exist in the extreme case when only one of your two sets is nonempty).