intersected topology
Let $X$ be a topological space with toplogy $\mathcal{T}$. If $Y$ is a subset of $X$, then we define the the toplogy $\mathcal{T}$ intersected with $Y$ as the set
${T}_{\cap Y}:=\left\{Y\cap U:U\in \mathcal{T}\right\}$
intersected topology is a topology
The set ${T}_{\cap Y}$ is a topology on $Y$

We'll show that $\mathrm{\varnothing }$ and $Y$ are elements of ${T}_{\cap Y}$, this can be seen as since $\mathrm{\varnothing }\in T$, then $Y\cap \mathrm{\varnothing }=\mathrm{\varnothing }\in {T}_{\cap Y}$ and also since $Y\subseteq X$, then $Y\cap X=Y$ so $Y\in {T}_{\cap Y}$

Now we want to prove that ${T}_{\cap Y}$ is closed under arbitrary unions, which we know because $\bigcup _{\alpha \in I}\left({U}_{\alpha }\cap Y\right)=\left(\bigcup _{\alpha \in I}{U}_{\alpha }\right)\cap Y$. Similarly we can see that it's closed under finite intersections as $\left({U}_{1}\cap Y\right)\cap \dots \cap \left({U}_{n}\cap Y\right)=\left({U}_{1}\cap \dots \cap {U}_{n}\right)\cap Y$.

Therefore we know that ${T}_{\cap Y}$ is a topology.

subspace topology
Since ${T}_{\cap Y}$ forms a topology, we will denote it by ${\mathcal{T}}_{\cap Y}$ and call it the subspace topology and that $Y$ is a subspace of $X$
basis for the subspace topology
Suppose that $\mathcal{B}$ is a basis for the topology of $X$, then $M:=\left\{Y\cap B:B\in \mathcal{B}\right\}$ is a basis and it generates ${\mathcal{T}}_{\cap Y}$

To prove this statement true, we'll use the handy basis criterion

First note that given any element $Y\cap B\in M$, we know that it is open with respect to ${\mathcal{T}}_{\cap Y}$, because $B$ is open with respect to $X$

Now let $Y\cap U$ be an open set of $Y$ which means that $U$ is open in $X$. Let $p\in Y\cap U$ , since $p\in U$ , $U$ is open in $X$ and $\mathcal{B}$ is a basis for $X$ then we know there is some $B\in \mathcal{B}$ such that $p\in B\subseteq U$ if that's the case then we know that $p\in Y\cap B\subseteq Y\cap U$, noting that $Y\cap B\in M$, therefore we know that $M$ is a basis and it is for $Y$

open in a subspace implies open
Suppose that $Y$ is a subspace of $X$ and also that $Y$ is open in $X$, then any set that's open in $Y$ is also open in $X$
Suppose that $U$ is open in $Y$, thus there is some $V$ open in $X$ such that $U=Y\cap V$, but since $Y$ is open in $X$, then as a finite intersection of open elements, then $U$ is also open in $X$, as needed.
TODO
Suppose that $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A×B$ is the same as the topology $A×B$ has as a subspace of $X×Y$

We will show that the two topologies are equal by showing that the basis that generate them are equal.

Given the subspace $A$ of $X$ and $B$ of $Y$, then the product topology of $A×B$ has a basis of elements of the form $\left(A\cap U\right)×\left(B\cap V\right)$ where $U,V$ are open in $X,Y$ respectively.

On the other hand ,an open set in $X×Y$ is of the form $U×V$ where $U,V$ are open in $X,Y$ respectively, thus a basis element for $A×B$ (as a subspace of $X×Y$)is a set of the form $\left(A×B\right)\cap \left(U×V\right)$.

Now to connect the two recall thqt $\left(A\cap U\right)×\left(B\cap V\right)=\left(A×B\right)\cap \left(U×V\right)$. So given a set, it is in the first basis if and only if it is in the second basis, meaning that the two bases are equal, and thus they generate the same topology.