Closed in X $\implies$ Closed in Y?

Let Y be an open subset of $\mathbb{R}^n$. If X is a closed subset of Y, disjoint from the boundary of Y, is it true that X is a closed subset of $\mathbb{R}^n$? How do I show this?

Edit: Let X be contained in a closed set B of $\mathbb{R}^n$ which is contained in Y and which is disjoint from the boundary of Y. Then X is closed in $\mathbb{R}^n.$

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There is a checkbox to the left of the answers. – AD. Sep 11 '12 at 19:38
@Vivek Ask the commenter to post the comment as a answer and accept it. – Ayman Hourieh Sep 11 '12 at 19:38
Maybe I'm misunderstanding the question, but how about $n=1$, $Y=(0,1)$, and $X=[1/2,1)$? You might ask the same question, but demand that the boundary of $Y$ be disjoint from the boundary of $X$. – David Mitra Sep 11 '12 at 19:54

If a set $A$ is not closed in $\mathbb{R}^n$, then there is a sequence $x_n\in A$ converging to $x\in \mathbb{R}^n\setminus A$. (Why?)