Closed sets in induced metric space 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 $Y\times Y$) 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. $q.e.d.$
 A: 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 $x\in X$ be a limit point of $E \subset X$. Then, $x \not\in E \implies x \not\in Y$.

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

Let $x\in X$ 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$}$
