Suppose $T$ a topological space and $S\subseteq T$ a subspace equipped with the subspace topology inherited from $T$. Take a subset $H\subseteq S$. I'd like to prove that $\partial_S H=S\cap \partial_T H$ (where $\partial_X A$ denotes the boundary of $A$ in $X$) but I suspect that this is true only if $S$ is open in $T$. First I've shown $\partial_S H\subseteq S\cap \partial_T H$ as follows: $x\in \partial_S H$ if and only if for every (open) neighbourhood $U\subseteq S$ of $x$ both $U\cap H\neq\emptyset$ and $U\cap H^c\neq\emptyset$ hold. Now, if $V$ is open in $T$ then $U=V\cap S$ is open in $S$ and $U\cap H, U\cap H^c\neq\emptyset$, so in particular $V\cap H, V\cap H^c\neq\emptyset$.
For the other inclusion, take $x\in S\cap\partial_T H$. Then, if $U=V\cap S$ is open in $S$ (where $V$ is open in $T$), we'd like to say that $U\cap H\neq\emptyset$ (similarly for $H^c$), but we should use that U is open also in $T$, which is true if $S$ is open in $T$.
Now, if this proof is correct, I am through for $S$ open in $T$. What can we say otherwise? Are the corresponding statements for the interior and the closure of $H$ still true? (I guess so, with an almost identical proof!)