Let $(X,\mathcal T)$ a topological space. Let $B\subset X$. Then the induced topology of $B$ is $$\mathcal T_B=\{U\cap B\mid U\in \mathcal T\}.$$ In other word, a set $A$ is open in $B$ if there is $U\in\mathcal T$ such that $$A=U\cap B.$$
But what is a closed set in $B$ ? I natural definition would be $A$ is closed in $B$ if $A^c$ is open in $B$. But the problem is that $A^c$ is not in $B$. Maybe $A$ is closed in $B$ if $A^c\cap B$ is closed in $B$ ? But I'm not really sure that it makes sense.
In every topology a set is closed iff it is the complement of an open set.
So in this case we find that sets of the form $B-(U\cap B)$ are closed.
Observe that $$B-(U\cap B)=B-U=U^{\complement}\cap B\tag1$$ where $U^{\complement}$ denotes the complement of $U$ as a subset of $X$.
So actually $(1)$ shows us that a set that is closed in $B$ can be written as $F\cap B$ where $F$ is a closed set in the original topological space $X$.
Also the converse of this is true: whenever $F$ is a closed set in $X$ then $F\cap B$ is a closed set in $B$.
This because $B-F\cap B=F^{\complement}\cap B$ where $F^{\complement}$ denotes the complement of $F$ in $X$, hence is open in $X$.