I'm reading this proof of the following claim.
Let $(X,d)$ be a metric space and $(Y,d')$ a subspace of $X$. If $(X,T)$ is the topology induced by $d$ and $(Y,T')$ the typology induced by $d'$, then $(Y,T')$ is a subspace of $(X,T)$.
Proof:
Since $d'$ is the restriction of $d$, an open ball in $(Y,d')$ is the restriction of an open ball in $(X,d)$ to $Y$. Consequently a subset $O'$ of $Y$ is open in $Y$ if and only if, for each $y\in O'$, there is an $\varepsilon>0$ such that $B(y,\varepsilon) \cap Y \subseteq O'$. Let $$O=\displaystyle \bigcup_{y\in O'}B(y;\varepsilon).$$ Then $O$ is open in $X$ and $O'=O\cap Y$. Thus $O'\in T'.$ Conversely, if $O'\in T'$ then $O'=O\cap Y$for some $O\in T$. For each $y\in O'$ we have $y\in O$ and $O$ is open, so there is an $\varepsilon$ such that $B(y,\varepsilon)\subseteq O$. It follows that $B(y,\varepsilon)\cap Y\subseteq O'$ and hence $O'$ is open in $Y$.
Now, to show that something is a subspace one has to show that every subset is open if and only if it's of the form $O\cap Y$--or at least, that's how I understand it from the definition in my text. It seems that the second half of the proof is showing the "if" part, but I don't see how the first half shows the "only if".
Letting $O$ be open in $T$ doesn't seem to be an assumption at any point in the proof. I'm sure I'm wrong about that somehow, but could anyone clarify where this assumption is made?
The proof starts with an arbitary $O \in T_{d'}$ (by which I mean the topology on $Y$ induced by $d'$, the restriction of $d$), and shows $O \in T'$ (where $T'$ is the subspace topology from $T$, the topology on $X$, which is induced by $d$). Then it takes any $O \in T'$, and shows it is in $T_{d'}$. So it shows the two inclusions, and $T' = T_{d'}$.
As $T'$ is by definition the subspace topology on $Y$, we are done showing that $d'$ ensures that $Y$ has the subspace topology.