Long proof of equivalence of subspace and metric topology Let $(X,d)$ be a metric space and $S\subseteq X$. Let $\tau$ be the topology on $X$ induced by $d$ and $\tau_S$ be the subspace topology on $S$:
$$
\tau_S = \{S\cap V:V\in \tau\}.
$$
Denote $B_r(x)= \{y\in X:d(x,y)<r\}$ and $B'_r(x)= \{y\in S:d(x,y)<r\}$.
I should prove that for $U\subseteq S$ it holds $U\in \tau_S$ iff for any $x\in U$ there is $r_x$ s.t. $B'_{r_x}(x)\subseteq U$.
Proof: Clearly, $B'_r(x) = S\cap B_r(x)$. 


*

*If $U\in \tau_S$ then there is $V\in \tau$ s.t. $U = V\cap S$. So, for any $x\in U\subseteq V$ there is $r_x$ s.t. $B_{r_x}(x)\subseteq V$ and hence $B'_{r_x}(x) = B_{r_x}(x)\cap S\subseteq V\cap S = U$.

*Conversely, if for any $x\in U$ there is $r_x$ s.t. $B'_{r_x}(x)\subseteq U$, then we can put
$$
V = \bigcup\limits_{x\in U}B_{r_x}(x)
$$
so $V\cap S =\bigcup\limits_{x\in U}B_{r_x}(x)\cap S = \bigcup\limits_{x\in U}B'_{r_x}(x) =U$.
If you write this proof formally that is quite long. I wonder if there are more direct and nice proofs.
 A: The indefinite articles in the second sentence are confusing. From the title and the rest of the question, it seems that what you actually mean is:

Let $\tau$ be the topology on $X$ (namely the one induced by $d$), and let $\tau_S$ be the subspace topology on $S$ (namely the one induced by $\tau$).

Another confusing aspect is that in part 2 you write "Conversely, if $U$ is open w.r.t. to the induced metric on $S$". This is of course equivalent to the actual premise ("for any $x\in U$ there is $r_x$ s.t. $B'_{r_x}(x)\subseteq U$"), but this equivalence is a significant part of what's to be proved. If you're already assuming this equivalence as known, then what you're trying to prove is what the title seems to suggest, namely the equivalence of the subspace topology $\tau_S$ on $S$ induced by $\tau$ and the metric topology $\tau'_S$ on $S$ induced by the metric on $S$ induced by $d$. In that case you can simplify the proof by not talking about the individual points of $U$ at all:
Equivalent claim: For $U\subseteq S$ we have $U\in \tau_S$ if and only if $U\in \tau'_S$.
Proof: Since the $B_r(x)$ form a base of $\tau$, the $B'_r(x)$ form a base of $\tau_S$: For $U\in\tau_S$, there is $V\in\tau$ with $U=S\cap V$ and $V=\bigcup B_{r_i}(x_i)$, so $U=S\cap\bigcup B_{r_i}(x_i)=\bigcup\left(S\cap B_{r_i}(x_i)\right)=\bigcup B'_{r_i}(x_i)$. By definition, the topology with base $B'_r(x)$ is $\tau'_S$.
[Edit regarding the clarified question:]
It's clear from the clarified question that you do want to start from the premise "for any $x\in U$ there is $r_x$ s.t. $B'_{r_x}(x)\subseteq U$" in part 2. In that case, you're not just proving what the title says, but the combination of that and "a set is open if and only if it is a neighbourhood for each of its points". The union of balls with one ball for each point is only required for that second claim, which has nothing to do with the equivalence of the subspace topology and the metric topology. Any proof of that second claim must necessarily use one ball for each point, since that's all that's given and the set wouldn't be open if there weren't such a ball for one of the points.
