I'm having trouble with this exercise (from Gamelin and Greene Introduction to Topology):
Prove that the closure $\bar{Y}$ of a subset $Y$ of a metric space $X$ coincides with the intersection of all closed subsets of $X$ that contain $Y$.
My thoughts so far: if we let $V$ be the intersection of all closed subsets of $X$ containing $Y$, then we want to show that $V \subseteq \bar{Y}$ and $\bar{Y} \subseteq V$. Clearly $V$ is closed (because it's the intersection of closed sets), and $Y \subseteq V$. We know from our definition that the closure of $Y$ consists of [edit: the following is wrong, as Brian and Levon pointed out] all points for which there exists a positive $r$ such that $B(x;r) \cap Y \neq \emptyset$. So effectively we're trying to show that $x \in V \Leftrightarrow \exists r B(x;r) \cap Y \neq \emptyset$.
And yet somehow I find myself at a loss as to how to proceed. Could someone please offer a hint? Or better, ideas on what key skills or knowledge I might be lacking if I'm finding such simple exercises unduly difficult? Thank you.
ADDENDUM--- Oh, dear, it was awfully careless of me to have confused the quantifier in the definition of closure; sorry about that.
SECOND ADDENDUM--- Okay, I think I have it now. The set $\bar{Y}$ is closed and contains $Y$, therefore it was one of the sets that we intersected to get $V$, and therefore $V \subseteq \bar{Y}$. It remains to show that $\bar{Y} \subseteq V$: that is, if a point belongs to $\bar{Y}$, then it therefore belongs to $V$. But by contraposition, this is the same as saying that if a point doesn't belong to $V$, then it doesn't belong to $\bar{Y}$. If a point $x$ doesn't belong to $V$, then there must be at least one closed set (call it $W$) containing $Y$ that does not contain $x$. Because $W$ is closed, its complement $X \backslash W$ (which contains $x$) is open. But that means there's an open ball $B(x;r)$ in $X \backslash W$. Then because $Y$ is not in $X \backslash W$, we can say that there exists an $r$ such that $B(x;r) \cap Y = \emptyset$. But by the De Morgan law, this is just what it means for $x$ not to belong to $\bar{Y}$, which is quod erat demonstrandum.