Gamelin Theorem 1.6 In Gamelin and Greene, Theorem 1.6 is stated:
Let Y be a subspace of a metric space X. Then a subset U of Y is open in Y if and only if $U = V  \cap  Y$ for some open subset V of X.
The proof starts:


*

*Because the metric d' : $Y \times Y \rightarrow R$ is the restriction to $Y \times Y$ of the metric d': $X \times X \rightarrow R$ on X, the open ball in the metric space $Y$ with center $y \in Y$ and radius $r > 0$ is just the intersection $B(y;r) \cap Y$ of $Y$ and the open ball $B(y;r)$ in $X$.  

*If $V$ is open in $X$, then for each $y \in V \cap Y$, there exists $r > 0$ such that $B(y;r) \subset V$.

*Then the open ball in $Y$ centered at $y$ with radius $r$ is contained in $V \cap Y = U$.

*Consequently each $y \in V \cap Y$ is an interior point of $V \cap Y$ in the subspace $Y$, so that $U = V \cap Y$ is open in Y.

I am confused by step 3. 
Does this theorem require that $V \subset Y$? 
Otherwise, I don't see that this step follows necessarily as part of the open ball B(y;r) might be outside of the intersection with Y. Perhaps someone can please clarify this for me?

It’s confusingly stated. What is meant in (3) is $B_{d'}(y;r)$, the open ball in the space $Y$, not $B_d(y;r)$, the open ball in the space $X$. The argument is that since $B_d(y;r)\subseteq V$ and $B_{d'}(y;r)=B_d(y;r)\cap Y\subseteq Y$, therefore $B_{d'}(y;r)\subseteq V\cap Y$.
