For $U\subseteq Y\subseteq X$, prove that $U$ is open in $Y$ iff there is a $V\subseteq X$ such that $U=Y\cap V$ Let $(X,d)$ be a metric space, with $Y$ a subset of $X$.

How do I prove that a subset $U\subseteq Y$ is open in the metric space $(Y,d|_{Y\times Y})$ iff there exists an open subset $V$ of $X$ such that $U=Y\cap V$?

So, I have to prove both sides. For the $"\implies"$-side, can I just take $V=U$? And the other side, I have no clue where to begin.
 A: No, because an open subset $U$ of $Y$ is not necessarily an open subset of $X$. Let's show that if $U \subseteq Y$ is open in $Y$ then we can find an open $V \subseteq X$ such that $U = V \cap Y$. By definition, for each point $p \in U$, we can find an open ball $B_Y(p,r_p)$ in Y such that $p \in B(p,r_p) \subseteq U$. Note that
$$ B_Y(p,r_p) = \{ q \in Y \,\, | \,\, d|_{Y \times Y}(p, q) < r_p \} = \{ q \in Y \, \, | \, \, d(p,q) < r_p \} = \{ q \in X \, \, | \,\, d(p,q) < r_p \} \cap Y = B_X(p,r_p) \cap Y $$
so the open ball in $Y$ around $p$ is the intersection of an open ball in $X$ around $p$ with $Y$. Let $V = \bigcup_{p \in U} B_X(p,r_p)$. Then $V$ is open in $X$ (as a union of open balls) and
$$ V \cap Y = \bigcup_{p \in U} B_X(p,r_p) \cap Y = \bigcup_{p \in U} B_Y(p,r_p) = U.$$
For the other direction, start with a point $p \in U = Y \cap V$. Since $p \in V$ and $V$ is open, we can find $r_p$ such that $p \in B_X(p,r_p) \subseteq V$. Then $B_Y(p,r_p) = B_X(p,r_p) \cap Y \subseteq V \cap Y = U$ is an open ball in $Y$ around $p$ that is contained in $U$.
