Let $Y$ be a subspace of $X$. Then $A\subseteq Y$ is closed in $Y$ if $A=C\cap Y$ for closed set $C$ in $X$ 
Let $Y$ be a subspace of $X$. Then $A\subseteq Y$ is closed in $Y$ if $A=C\cap Y$ for closed set $C$ in $X$

Since $A$ is closed in $Y$, then $Y\setminus A$ is open in $Y$.
Then $Y\setminus A = U\cap Y$ for some open set $U$ in $X$.
Then $U^c$ is a closed set in $X$.
I want to say $A=U^c\cap Y$ but I'm not sure how.
 A: Here's some general advice:


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*If you are trying to show that two sets $T$ and $S$ are equal, it is often helpful to do it by checking both $T \subset S$ and $S \subset T$.

*For any two sets $T$ and $S$, the definition of $T \setminus S$ is $T \cap S^c$.

*For any two sets $T$ and $S$, $(T \cap S)^c = T^c \cup S^c$.


We don't actually need 1. but it's a good piece of general advice!
So in your case you have already said that $Y \setminus A = U \cap Y$. This means that $Y \cap A^c = U \cap Y$. By taking complements, this implies that $Y^c \cup A = U^c \cup Y^c$. Now intersect both sides with $Y$. You get $A \cap Y = U^c \cap Y$. But $A \subset Y$, so $A \cap Y = A$. So we've shown $A = U^c \cap Y$.
A: So we know $A \subseteq Y\tag{1}$ and $Y \setminus A= U \cap Y\tag{2}$
We want to show $U^\complement \cap Y = A\tag{3}$
So let $x \in A$, then for sure $x \in Y$ and if $x \in U$ would hold, $x \in Y\setminus A$ by $(2)$ so $x \notin A$, contradiction. So $x \notin U$ and hence $x \in Y \cap U^\complement$ and we have shown one inclusion of $(3)$. Conversely, if $x \in Y \cap U^\complement$ then $x \in Y$ and $x \notin U$. So $x \notin Y \setminus A$ and this can happen only if $x \in A$ and so the other inclusion is also clear.
In case of doubt try the two inclusion method...
