$U \in \tau(\Bbb R,d_\Bbb R)$ $\iff$ there are open intervals $B_1,B_2,B_3,...$ with $U= \bigcup_{n\in\Bbb N} B_n$ 
The rational numbers are countable: you can write $\Bbb Q =${$q_1,q_2,q_3,...$}.
Moreover,$\Bbb Q$ is dense in $(\Bbb R,d_\Bbb R)$.
Use these facts to prove for a non-empty set  $U\subseteq \Bbb R$, we have:
$U \in \tau(\Bbb R,d_\Bbb R) \iff$ there are open intervals $B_1,B_2,B_3,...$ with $U= \bigcup_{n\in\Bbb N} B_n$


Definitions: 


*

*Let $(X,d)$ be a metric space. We say that a set $D\subseteq X$ is dense in $(X,d)$  If$\,$ $\overline  D =X$

*$\tau(X,d)=${$U\subseteq X : U $ is open in $(X,d)$} [the topology of a metric space]

 A: Outline: given $x \in U$ find an open interval $(x-r,x+r)$ about $x$ that is contained in $U$. Since the rational numbers are dense in the reals, there exists a rational $q_i$ such that $|x-q_i| < r$. Use this to find an open interval about $q_i$ that (a) is contained in $(x-r,x+r)$, and (b) contains $x$.
A: SKETCH: If $U$ is open, and $x\in U$, then by definition there is some open interval $(a,b)$ such that $x\in(a,b)\subseteq U$. (If your definition of open set requires that $(a,b)$ have the form $(x-\epsilon,x+\epsilon)$ for some $\epsilon>0$, that’s fine: it doesn’t affect the argument.) Use the fact that $\Bbb Q$ is dense in $\Bbb R$ to argue that there are rational numbers $p\in(a,x)$ and $q\in(x,b)$. Let $$\mathscr{B}_U=\big\{(p,q):p,q\in\Bbb Q\text{ and }p<q\text{ and }(p,q)\subseteq U\big\}\;.$$ Show that $\mathscr{B}_U$ is countable and that $\bigcup\mathscr{B}_U=U$.
In fact with more work one can prove that $U$ is the union of an at most countable collection of pairwise disjoint open intervals; see this question and the answers to it.
