Existence of open ball in whole space from the existence of nonempty open set in subspace

Question

I am reading some mathematical article but I don't understand one step in the proof. Assume that $(X,d)$ is a complete and separable metric space (maybe this is not important that the space is complete and separable, but I'm not sure) and assume that $A\subset X$ is open and nonempty. Define $Y:=cl A$ (closure of A in $(X,d)$) as a subspace of the space $(X,d)$ and assume that there exist a set $V\subset Y$ open in the space $Y$. We have to prove that there exist an open ball $B$ in the space $(X,d)$ such that $B\subset V$

Here you can find more details of my question. We know also that $$Y=\cup_{n=1}^{\infty}Y_n$$ where $Y_n$ are closed in $Y$ and we know that there exist $N$ such that $int Y_N\neq \emptyset$ (interior in subspace $Y$) Then there exist a ball $B$ in the space $(X,d)$ such that $B\subset Y_N$. This is the right question.

Answer 1

Since $V$ is an open of $Y$, it can be written $Y \cap U$ for an open $U$ of $X$. Then $U \cap A$ is an open of $X$ (non empty is $V$ is not empty), so contains a ball $B$ and finally $B \subset A \cap U \subset Y \cap U=V$.

Answer 2

The set $V$ is irrelevant ! Take $a \in A$. Since $A$ is open, there is an open ball $B$ such that $a \in B \subset A$. Hence $B \subset cl(A)=Y.$

Answer 3

Is this not the definition of an open set ?

By definition, $\mathrm{Int}(Y_n)$ is an open set : if it is non-empty, there is a point $a \in \mathrm{Int}(Y_n)$, and by definition of an open set, you can find a ball (that contains $a$) contained in $\mathrm{Int}(Y_n)$, so contained in $Y_n$.