Open set that is the intersection of sets in a metric space Let $(X,d)$ be a metric space ;and $Y \subset X$. Suppose $G \subset X$ is open in $X$ .Show that $G \cap Y$ is open in $(Y,d)$.
My attempt: clearly $G \cap Y \subset Y$. If $G \cap Y=\emptyset$ then we know that $\emptyset$ is open in $(Y,d)$. Suppose that $G\cap Y \ne \emptyset$ then for any $z \in G \cap Y$ we have that $z \in G$ so there is an $\epsilon>0$ such that $B(z: \epsilon) \subset G$, similarly, since every metric space is open, there is an $\epsilon_{1}>0$ such that $B(z:\epsilon_{1}) \subset Y$. Now define $r_{z}=\min(\epsilon,\epsilon_{1})$ then for any $x \in B(z:r_{z})$ we have $d(z,x)< \epsilon$ so $x \in G$ and $B(z:r_{z}) \subset G$ , also $d(z,x)<\epsilon_{1}$ so $x \in Y$ and $B(z:r) \subset Y$ . Hence we have that $B(z:r_{z}) \subset (G \cap Y)$ , since it holds for every $z$ ; $$G\cap Y=\bigcup_{z \in G \cap Y} B(z:r_{z})$$ which is an union of open sets in $Y$ which in turn is open in $Y$ by proposition 1.9(at least in this book).
The solutions manual offers another proof, so I wondered If this proof correct?
This is exercise 8 chapter 2 of "Functions of One Complex Variable 1- John B Conway"
 A: Hint: Be more explicit in what $B(z,r_z)$ 
The idea of the proof is correct. However, I would suggest you to be more explicit and differentiate between what is an open ball in $X$ and what is an open ball in $Y$. 
You wrote that "every metric space is open", which is technically not true. Openness is a property, relative to a topology i.e. a set is not open, it is open with respect to (w.r.t.) a topology $\tau$. For metric spaces this means the following: 
A set $A \subseteq X$ is open in a metric space $(X,d)$ if $\forall z \in A, \exists \varepsilon > 0: s.t. B{(z, \varepsilon)} \subseteq A$, where $B{(z, \varepsilon)} := \{a \in X \vert d(z,a) < \varepsilon\}$. Note that the elements of $B{(z, \varepsilon)}$ are in $X$.
Now in your proof you should keep track with repect to which metric space an open ball is defined and maybe write $B^X{(z, \varepsilon)} := \{a \in X \vert d(z,a) < \varepsilon\}$ for the open balls in $X$ and $B^Y{(z, \varepsilon)} := \{a \in Y \vert d(z,a) < \varepsilon\}$ for the open balls in $Y$. Then as you write above with $r_z := \min(\varepsilon, \varepsilon_1)$ we have $B^Y{(z,r_z)} \subseteq B^X{(z,\varepsilon)} \subseteq G$ and $B^Y{(z,r_z)} \subseteq B^Y{(z,\varepsilon_1)} \subseteq Y$. Hence $B^Y{(z,r_z)}$ is an open ball in $Y$ with center $z$ contained in $G \cap Y$. Since $z \in G \cap Y$ was arbitrary, this makes $G \cap Y$ open in $Y$.

It is nessecary to define $B^Y{(z,r_z)}$ as an open ball in $Y$. Otherise the argument is false. Consider $X = \mathbb{R}$ with the usual metric and $Y= [0,1]$. Then for $0$ there does not exists an $\varepsilon > 0$ s.t. $B^X{(0,\varepsilon)} \subseteq Y$ (making $Y$ not open in $X$). However, $[0,1]$ is of course open in $Y$ since by definition, for any $\varepsilon > 0$ we have $B^Y{(0,\varepsilon)} \subseteq Y$.
