Check my answer: Prove that every open set in $\Bbb R^n$ is a countable union of open intervals. I have a question. I have solved this but please can you check my solution? Thank you. 
If there are any mistakes or something is missing and so on, please tell me. 
This is important to me. Is this proof enough to get a successful grade on an exam? 
Btw, I underlined the question with pink a pencil. 

 A: There are two problems with your argument. The first is when you have 
$$V\subseteq\bigcup_{x\in A_0}B_{\epsilon_x}(x)\;,\tag{1}$$
where $A_0$ is a countable subset of $X$, and say that $A_0=\Bbb N$. $A_0$ is not $\Bbb N$: it’s a subset of $X$. If it’s a countably infinite subset of $X$, then there is a bijection from $\Bbb N$ to $A_0$, and you can enumerate $A_0=\{x_j:j\in\Bbb N\}$, let $B_j(x_j)=B_{\epsilon_j}(x_j)$ for each $j\in\Bbb N$, and say (as you did) that
$$V\subseteq\bigcup_{j\in\Bbb N}B_j(x_j)\;,\tag{2}$$
but you have to say that that is what you’re doing. However, $A_0$ might be finite, in which case there is no bijection from $\Bbb N$ to $A_0$.
There’s no need to do all of this, however: $(2)$ is an unnecessary rewriting of $(1)$ even when $(2)$ is correct. If $A_0$ is countable, then the union is $(1)$ is a countable union, and that’s all you need. However, I would strengthen $(1)$ and say that
$$V=\bigcup_{x\in A_0}B_{\epsilon_x}(x)\;:\tag{3}$$
you’ve actually proved this, and it’s what you need: $V$ is a countable union of open balls, not just a subset of a countable union of open balls.
Now we come to the larger problem. Your last step does not work at all: a ball in $\Bbb R^n$ is not an interval. To complete your proof, you need to show that an open ball in $\Bbb R^n$ is a countable union of intervals. Then you can argue like this: 

For each $x\in A_0$ there is a countable family $\mathscr{I}_x$ of intervals in $\Bbb R^n$ such that $B_{\epsilon_x}(x)=\bigcup\mathscr{I}_x$. Let $\mathscr{I}=\bigcup_{x\in A_0}\mathscr{I}_x$; then $\mathscr{I}$, being the countable union of countable sets, is countable, and $$B=\bigcup_{x\in A_0}\bigcup\mathscr{I}_x=\bigcup\mathscr{I}$$ is a countable union of intervals.

To do this, though, you have to prove that if $B_r(x)$ is any open ball in $\Bbb R^n$, then $B_r(x)$ is the union of countably many sets of the form
$$[a_1,b_1)\times[a_2,b_2)\times\ldots\times[a_n,b_n)\;.$$
I suggest the following approach.


*

*First prove that $B_r(x)$ is the union of countably many sets of the form $$(a_1,b_1)\times(a_2,b_2)\times\ldots\times(a_n,b_n)\;.$$ You can do this by showing that if $y\in B_r(x)$, there are rational numbers $p_1,\dots,p_n,q_1,\dots,q_n$ such that $$y\in(p_1,q_1)\times(p_2,q_2)\times\ldots\times(p_n,q_n)\;.\tag{4}$$ There are only countably many rational numbers, so there are only countably many open boxes like $(4)$.

*Show that each open box like $(4)$ is the union of countably many intervals. HINT: Use the fact that in $\Bbb R$, $$[a,b)=\bigcup_{n\in\Bbb Z^+}\left(a+\frac1n,b\right)\;.$$
