Closed subsets in metric space I want to prove that any closed subset $F$ from a metric space $(E,d)$ can be written as a denumerable  intersection of open sets i.e., 
$$F=\bigcap_{n\in\mathbb{N}} \Theta_n; \Theta_n=\bigcup_{x\in F} B(x,\frac{1}{n})$$
It is clear that $F\subset \bigcap_{n\in\mathbb{N}} \Theta_n$ 
I suppose that $y\in \Theta_n$ for all $n\in \mathbb{N}^*$ and $y\notin F$ i.e., $\exists \delta>0; B(y,\delta)\cap F\neq \emptyset$ but the fact $y\in \Theta_n$ means that there exist $x\in F$ such that $d(x,y)<\frac1n, \forall n\in \mathbb{N}$ it still right for $n$ such that $\frac1n<\delta$ and then $x\in F\cap B(y,\delta)$ contradiction.
But I don't know where to use the fact that $F$ is closed and $\Theta_n$ is open? 
Thank you.
 A: Hint: For every subset $S$  define $$d(x,S)=\inf\{d(x,s):s\in S\} $$ which is a continous function on $E$ 
and try to show that $$F=\bigcap_{n\in\mathbb{N}} \theta_n$$ where $$\theta_n =\{x\in E : d(x,F)<\frac{1}{n} \}$$
because of continuity $\theta_n$ is open for every $n\in \mathbb N$
A: Suppose $y\in\Theta_n$, for all $n$ and $y\notin F$.
Since $F$ is closed, there exists $\delta>0$ such that $B(y,\delta)\cap F=\emptyset$.
Consider $n$ such that $1/n<\delta$. Since $y\in\Theta_n$, there is $x\in F$ such that $y\in B(x,1/n)$ (by definition of $\Theta_n$).
Then $d(x,y)<1/n<\delta$, so $x\in B(y,\delta)$. Since $x\in F$, we have $x\in B(y,\delta)\cap F$, so $B(y,\delta)\cap F\ne\emptyset$.
Contradiction.
Therefore such a $y$ must belong to $F$ and we have proved that
$$
F=\bigcap_{n>0}\Theta_n
$$
Since $\Theta_n$ is open by construction (a union of open spheres), it follows that every closed set in a metric space is a countable intersection of open sets (a $G_\delta$, in a common terminology).

Addition
An alternative way of proving that, if $y\in\Theta_n$ for all $n$, then $y\in F$ is as follows.
Let $n>0$; then $y\in \Theta_n$, so $y\in B(x_n,1/n)$, for some $x_n\in F$. Let's see that the sequence $(x_n)$ converges to $y$.
Fix $\varepsilon>0$ and take $n$ such that $n>1/\varepsilon$. Then
$$
d(x_n,y)<1/n<\varepsilon
$$
Thus, taking $\bar{n}>1/\varepsilon$, we see that, for every $n\ge \bar{n}$, we have $d(x_n,y)<\varepsilon$, which amounts to our thesis.
Since $F$ is closed and $y$ is the limit of a sequence in $F$, then $y\in F$.
Note. This is essentially the same as the proof by contradiction above. It's neither more nor less correct.
