I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Suppose $Y \subset X$, where $X$ and $Y$ are metric spaces. Prove that a subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.
My Proof:
Suppose $E$ is open relative to $Y$. Then, for each $e \in E$ there is some $r$ s.t. $B_r(e) \cap Y \subset E$ where $B$ is an open ball in $X$. Now, let $G$ be the union of all $B_r(e)$ for every $e \in E$, and so $G \cap Y \subset E$. We can see that $B_r(e)$ is an open subset of $X$, and therefore $G$ is a collection of open subsets of $X$ and is thus an open subset of $X$. So $Y \cap G \subset E$ for an open subset $G$ of $X$. For any $e \in E$, $e \in B_r(e)$ for its associated open ball and since $E \subset Y \subset X$, $e \in Y$ and $e \in X$. So $e \in B_r(e) \cap X = G_e$ and so $e \in G$ and $E \subset G$. Thus, $E \subset Y \cap G$ and so $E = Y \cap G$.
Now suppose $E = Y \cap G$ for some open subset $G$ of $X$. For every $e \in E$ there is some $r$ such that $q \in G$ if $d(p,q) < r$. If $q$ is also an element of $Y$, then $q \in Y \cap G$. So, for every $e \in E$ there is some $r$ such that $q \in Y \cap G = E$ if $d(p,q) < r$ and $q \in Y$, so $E$ is open relative to $Y$.
Best Answer
The proof is correct, but it seems a bit too wordy to me. I'd shorten the first part of your proof to
Since this part makes use of open balls, I would prefer to use the same idea for the converse as well, e.g.