Ytukyg

Let $(X,d)$ be a metric space, let $Y$ be a subset of $X$ and let $E$ be a subset of $Y$, then

$(i)$ $E$ is relatively open with respect to $Y$ if and only if $E=V cap Y$ for some set $V subseteq X$ which is open in $X$.

$(ii)$ $E$ is relatively closed with respect to $Y$ if and only if $E=K cap Y$ for some set $k subseteq X$ which is closed in $X$.

I need to prove $(ii)$. I looked at this solution for a similar problem in a topological space, but it uses $(i)$. I tried to write a proof that doesn't depend on (i), but I don't know if it works. I am particularly suspicious about the bold text:

$Proof.$ Let $E$ be closed in $(Y,d)$ (with the metric $d$ restricted to $Ytimes Y$) and let $K = bar E$, where $bar E$ is the closure of $E$ with respect to $(X,d)$. Then $K$ is closed in $(X,d)$ and since $K$ is the union of adherent points to $E$ in $(Y^c,d)$ and adherent points to $E$ in $(Y,d)$, then $K cap Y =$ adherent points to $E$ in $(Y,d)$. Since $E$ is closed, it contains all its adherent points, so $K cap Y = E$.
Similarly, let $K$ be a subset of $X$ closed in $(X,d)$ such that $K cap Y = E$ , then $E$ = all adherent points of $K$ in $Y$. Thus $E$ is closed. $q.e.d.$

Your proof is correct!

Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.

First, observe that if $x$ is a limit point of a set $A$ in $Y subset X$, then $x$ is also a limit point of $A$ in $X$.

Now, let $K subset X$ be the closure of $E$ in $X$, as you have defined it. We want to show that $K cap Y = E$; that is, if $x$ is a limit point of $E$ in $X$ that does not lie in $E$, then in fact $x$ does not lie in $Y$. In other words, we want to prove the following:

Let $xin X$ be a limit point of $E subset X$. Then, $x notin E implies x notin Y$.

But now, the contrapositive of this statement makes things clear:

Let $xin X$ be a limit point of $E subset X$. Then, $x in Y implies x in E$.

But, this is true simply because, as we observed, if $x in X$ is a limit point of $E$ and $x$ also belongs to $Y$, then $x$ is a limit point of $E$ in $Y$. Since $E$ is closed in $Y$, this implies that $x in E$. $tag{$blacksquare$}$

The proof seems fine to me..
E is closed so must contain all points $ x in Y $ such that for some $ epsilon > 0$, $ B_{epsilon}(x) cap E neq emptyset $, and these are exactly the points in Y that also belong to the closure K.