Proof of "closed set is G-Delta set". Where did I make mistakes in this proof? I am studying Lebesgue integral and Lebesgue measure theory and I have come across one proportion.
I want to prove that all closed sets of $\mathbb{R^n}$ are $G_{\delta}$ set.
Here is the proof but I think that I made mistakes.
Let $F \subset \mathbb{R^n}$  a closed set and $F_{\frac{1}{n}} := \cup_{x\in F} B\bigg(x, \dfrac{1}{n}\bigg)$, where $B\bigg(x, \dfrac{1}{n}\bigg)$ is a open ball around $x$ with radius $\dfrac{1}{n}.$
Now, because $F \subset F_{\frac{1}{n}}$ for all $n \in \mathbb{N},$ $F \subset \cap_{n=1}^{\infty} F_{\frac{1}{n}}$.
I'll prove that $F \supset \cap_{n=1}^{\infty} F_{\frac{1}{n}}$.
Let $y \in \cap_{n=1}^{\infty} F_{\frac{1}{n}}$ and $\epsilon>0$  positive.
From Archimedes' principle, there exists $N\in \mathbb{N}$ such that $\dfrac{1}{N}<\epsilon.$
Because $y \in F_{\frac{1}{N}},$ $y \in \cup_{x\in F} B \bigg( x, \dfrac{1}{N} \bigg)$.
So there exists $x_0 \in F$ such that $y \in  B \bigg( x_0, \dfrac{1}{N} \bigg)$.
Now, $d(x_0, y) < \dfrac{1}{N} < \epsilon$ and from the arbitrariness of $\epsilon,$ $d(x_0, y)=0$ thus $y=x_0 \in F$.
So $F \supset \cap_{n=1}^{\infty} F_{\frac{1}{n}}$ and $F$ is $G-$Delta set. $\ \rule[-1mm]{2mm}{2mm}$
But this proof doesn't use the fact that $F$ is closed.
I think that this proof has some sort of mistakes.
Where did I mistake?
 A: You made your mistake when you concluded that $d(x_0,y)=0$. In fact all that you know is that $d(x_0,y)<\frac1N$. If you now choose an $\epsilon'<\epsilon$ needing an $N'>N$ in order to have $\frac1{N'}<\epsilon'$, it’s entirely possible that $y\notin B\left(x_0,\frac1{N'}\right)$: there must be an $x_1\in F$ such that $y\in B\left(x_1,\frac1{N'}\right)$, but it need not be $x_0$.
What you can do instead is note that for each $n\in\Bbb Z^+$ there is an $x_n\in F$ such that $y\in B\left(x_n,\frac1n\right)$, and therefore $x_n\in B\left(y,\frac1n\right)$. The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ is then a sequence in $F$ converging to $y$, and $F$ is closed, so $y\in F$.
An alternative approach is to show that if $y\notin F$, then $y\notin\bigcap_{n\in\Bbb Z^+}F_{1/n}$. Since $F$ is closed, $y\notin F$ implies that there is a $k\in\Bbb N$ such that $B\left(y,\frac1k\right)\cap F=\varnothing$. But then $y\notin F_{1/k}$, so $y\notin\bigcap_{n\in\Bbb Z^+}F_{1/n}$. It follows that $\bigcap_{n\in\Bbb Z^+}F_{1/n}\subseteq F$.
