Let's say that $(y,d')$ is a metric subspace of $(x,d)$. My professor just wrote a bunch of properties that can help finding the interior, closure and etc relationship between the two spaces. I could prove all of them but I couldn't manage to prove that the set of accumulation points in $Y$ is a subset of the set of accumulation points in $X$ I tried to take some elements in the left side as usual but I couldn't manage.
$x \in A'$ means that for every $r>0$ $(B_r(x)-\{x\}) \cap A \neq \emptyset$ And I know that any neighborhood in $Y$ is just a subset of the neighborhood in $X$
As you have said. Take $x\in A_Y'\subset Y$, then for every ball $(B_y(x,r)\backslash \{x\})\cap A\neq \emptyset$ in $Y$. Now take an open ball in X $B_X(x,r)$, then $B_X(x,r)\cap Y$ is an open set in Y, thus $((B_x(x,r)\cap Y)\backslash \{x\}))\neq \emptyset$, so $x\in A'_X\subset X$.
In case that you need here is one example. Take $X=[0,1]$, $Y=[0,1)$ and $A=(0,1)$ with the euclidean metric. Then the accumulation points of $A$ in both cases are the total space $X$ and $Y$, but they are different.