If $U\subseteq X$ is open in $X$, then $U\cap Y$ is open in $Y$ If $X$ is a metric space, and $Y$ is a metric subspace of $X$ the show that if $U\subseteq X$ is open in $X$, then $U\cap Y$ is open in $Y$.
So we have two cases: if $U\cap Y=\varnothing$ and $U\cap Y\neq\varnothing$. The first one is trivial (and is open).
Let's say that $U\cap Y\neq\varnothing$. Because $U$ is open, then for every $u\in U$,  $\exists \delta>0$ such that $B_X(u,\delta)\subseteq U$, in particular if we consider only the $u$'s in $U\cap Y$ then exists some $\varepsilon>0$ such that $B_X(u,\varepsilon)\subseteq U\cap Y$. Now because $X$ and $Y$ have the same metric, then if $u\in U\cap Y$ then $B_X(u,\varepsilon)=B_Y(u,\varepsilon)$, and we're done.
Now, I don't know if this is enterly correct, sincerely I think is not, but I'm stuck; a friend told me that it was easier to use that continuous functions send open sets to open sets, but I don't know how to use that for this problem.
 A: For metric spaces $T$, $O \subseteq T$ is open if for all $x \in O$, there exists an $\epsilon$ such that $B_T(x, \epsilon) \subseteq U$. 
Now let $X$ and $Y$ be as in your question. Note that for all $\epsilon$ and $x \in Y$, $B_Y(x, \epsilon) = B_X(x, \epsilon) \cap Y$ since the metric on $Y$ is the just the metric on $X$ restricted to $Y$. 
So let $U$ be open in $X$. Let $x \in U \cap Y$. In particular $x \in U$. So since $U$ is open in $X$, there is an $\epsilon$ such that $B_X(x, \epsilon) \subset U$. Then $B_Y(x, \epsilon) = B_x(x,\epsilon) \cap Y \subseteq U \cap Y$. Hence it has been shown that for all $x \in U \cap Y$, there is an $\epsilon$ such that $B_Y(x, \epsilon) \subset U \cap Y$. So $U \cap Y$ is open in $Y$. 

If you are familiar with point set topology: This amounts to saying the induced metric topology on $Y$ is the subspace topology induced by the metric topology on $X$. 
A: Y is open, U is open, so their intersection is open.
It is not true, btw, that continuous functions send open sets into open sets.  Consider f(x) = x^2 on (-1,1).
It is true that if f is continuous, then f(X) open implies X open.  In a metric space, this is equivalent to the usual epsilon-delta definition.
