Hi I encountered this problem in Analysis 1 from Otto Forster. I know that there exist answers already concerning this question, but I came up with a proof and am not sure whether the proof is valid or not, if it is not valid, I'm very curious as to why is it not valid. The problem is stated below:
A subset $U\subset\mathbb{R}$ is called open, if to every $a\in U$ there is an $\epsilon>0$ so that
$$]a-\epsilon,a+\epsilon[ \subset U $$
And we were asked to show that every open subset $U\subset\mathbb{R}$ is an union of countably many open intervals.
The definition that this book uses: For any set $A$, $A$ is called countable if there exists a surjection from $\mathbb{N}$ to $A$
I argued:
$U$ must be an union of open intervals, suppose it is not, then $U$ must contain some $u$ that there exists no $\epsilon > 0$ so that $]u-\epsilon,u+\epsilon[\subset U $, a contradiction to the definition of $U$.
First we show that for any open subset $U$, we can describe it with non-overlapping open intervals, in other words, it is the union of open intervals but none of them overlaps.
Suppose two open intervals $i=]a-\epsilon,a+\epsilon[$ and $j=]b-\epsilon',b+\epsilon'[$ overlap, w.l.o.g $a<b$,if they overlap "partly", we have $$a-\epsilon<a<b-\epsilon'<a+\epsilon<b<b+\epsilon'$$
we can then form a new interval by taking their midpoint. Also if $j$ is included within $i$, we can just throw away $j$.
Suppose $U$ is not empty, if it's empty then the statement holds vacuously. Then we pick some $u\in U$. So now let $U$ consists of non-overlapping open intervals, we can then pick an unique open interval $T$ so that $u\in T$, now we define $F$ to be the set of the intervals and a function $f:F\longrightarrow \mathbb{Q}$ with
$$f(S) = x,\;\;x\;is\;the\; rational\; number\; with \;the \;smallest\; dividend\;within \;S $$
Since the intervals are non-overlapping, $f$ is an injection.
Now we define $g:\mathbb{Q}\longrightarrow F$ as following: $$g(x) = S\;\;\;\; \;if\; f(S)=x$$
otherwise, $$g(x) = T$$
So that $g$ is a surjection (onto). Since $\mathbb{Q}$ is countable, let $h:\mathbb{N}\longrightarrow\mathbb{Q}$ be a surjection, so that $g \circ h:\mathbb{N}\longrightarrow F$ is also a surjection. Therefore $F$ is countable.
Thanks!