Proving a set open in a subset of a metric space is open in the whole space I have a few questions about this proof:

let $(X,d)$ be a metric space. Let $Y \subseteq X$ be an open subset
  in $X$. If $G \subseteq Y$ is open in $Y$, prove that $G $ is open in
  $X$.
PF: Since $G$ is open in $Y$, there is an open set $U$ in $X$ such
  that $G = U \cap Y$. Since $Y$ is open in $X$, it follows that $G$, as
  an intersection of two open sets in $X$, is open in $X$.

How do we know that $U$ exists? Why do we need to intersect it with $Y$?
 A: When that says "$G \subseteq Y$ is open", I believe it's intending for us to think of $Y$ itself as a topological space endowed with the subspace topology.  Given any topological space $X$ and any subset $Y \subset X$, the subspace topology on $Y$ is defined such that the open sets are $\{U \cap Y \ | \ U \text{ is open in } X \}$.  As an exercise, try proving that this is a legitimate topology.
Now, when $Y$ is an open subset of $X$ as it is here, we need not get too hung up on this formality because any $U \cap Y$ is open both in $X$ and in $Y$ under the subspace topology.  This is to say, the open subsets of $Y$ are simply the opens of $X$ that happen to be contained in $Y$.  However, we need to be more careful when $Y$ is not open in $X$.  For instance, $(0, 1]$ is open in $[-1,1]$ under the subspace topology inherited from $\mathbb{R}$ since $(0,1] = (0,2) \cap [-1,1]$, but $(0,1]$ is not itself open in $\mathbb{R}$.
A: For general topological spaces, that is precisely the definition of being open in a subspace. For metric spaces, we can spell it out:
Being open in $Y$, $G$ is the union of open balls (in $Y$), i.e., 
$$ G=\bigcup_{i\in I}B'(x_i,r_i)$$
with $x_i\in Y$, $r_i>0$, and $B'(x,r)=\{\,y\in Y\mid d(x,y)<r\,\}$.
This makes $B'(x,r)=Y\cap B(x,y)$, where $B(x,r)=\{\,y\in X\mid d(x,y)<r\,\}$ is an open ball in the larger metric space $X$. Thus
$$ G=\bigcup_{i\in I}(Y\cap B(x_i,r_i))=Y\cap \underbrace{\bigcup_{i\in I}B(x_i,r_i)}_{=:U}$$
