So, I'm trying to prove that if I have $A,B,C$ such that $C \subset B \subset A$, and $\rho$ is a metric in $A$, and $C$ is open in $(A,\rho)$, then $C$ is open in $(B, \rho\big|_{B})$.
In general, I want to see what happens if $C$ is $i)$ open in $A$, $ii)$ closed in $A$, $iii)$ open in $B$ or $iv)$ closed in $B$.
For the last case, I can prove that:
if $C$ is closed in $(B,\rho\big|_{B})$, then $C$ is closed in $(A,\rho)$: Since being closed implies $B-C$ is open, then for any element $c \in B-C$, there exist an open ball $B_{B}(c,r)$ (with $r>0$) such that $B_{B}(c,r) \subset B-C$, then since $B\subset A$, $B-C\subset A-C$. Then, $B_{B}(c,r) \subset A-C$ and that way $A-C$ is open, and $C$ is closed in $A$.
Now, for $i)$, I believe the proof is something like this:
if $C$ is open in $A$, then for any $c\in C$, there exist some $r_{c}>0$ such that $B_{A}(c,r_c) \subset C$. Then, since $C \subset B$, $C \cap B = C$, then $C$ is open in $B$, since now consider $B_{A}(c,r_c) \cap B = B_{B}(c,r'_c)$. Then $(B_{A}(c,r_c) \cap B) \subset (C \cap B)$, this is, $B_{B}(c,r'_c) \subset C$
Are these two proofs correct?
Any hint about $ii)$ and $iii)$ would be greatly appreciated.
Closed in $B$ does not imply closed in $A.$ Example $A=\Bbb R,B=C=(0,1)$ (for the usual distance). So your first proof cannot be correct.
Similarly, open in $B$ does not imply open in $A.$ Example $A=\Bbb R,B=C=[0,1].$
Your second proof is correct but this theorem can be easily proved more generally for an arbitrary topological space $A$ (not necessarily metrizable) and the induced topology on a subset $B$ (the open subsets of $B$ are by definition the $U∩B$ for $U$ open subset of $A$).
Similarly, in the same context, if $C$ is closed in $A$ then it is closed in $B.$