Prove directly that the space $\mathbb{R}$ of real number with usual absolute value metric is second countable. I can do this by proving $B=\{N_q (x):$ $x$ and $q$ are rational $\}$ form  a countable basis.
Since $x,q \in \mathbb Q$, and rational is countable so the above set is countable.
Since the union of members of $B$ is $X$ and intersection of members of $B$ is also  a member of $B$ since $B$ is finite and finite intersection of open sets is open. 
Let $N_1,N_2∈B$, then  $N_1=N_{1/a} (x)$ and $N_2=N_{1/b} (x)$ . Suppose $a≤b$, then $1/a≤1/b$, so $N_{1/a} (x)⊆N_{1/b} (x)$, thus  $N_1∩N_2=N_1∈B$
So $B$ is a basis.
I'm not sure if this argument solid enough or making sense?
 A: To finish your argument it suffices to express any ball $N_r (x)$ as a union of balls from $B$ (for arbitrary $x,r \in \mathbb{R}$ with $r>0$. To do this just write
$$ N_r (x) = \bigcup_{q \in N_r (x) \cap \mathbb{Q}} N_{\alpha(q)}(q) $$
where $\alpha(q)$ is any rational number with the property that $N_{\alpha(q)}(q) \subset N_r (x)$.
A: Your argument so far is fine, but you are not quite done. Since every element of $\mathcal B$ is open in the usual topology on the real line, then to show that $B$ is a basis for said topology, you should show that (1) for any open $U\subseteq\Bbb R$ and any $y\in U,$ there is some $V\in\mathcal B$ such that $x\in V\subseteq U,$ and (2) for any $V_1,V_2\in\mathcal B$ and any $z\in V_1\cap V_2,$ there is some $W\in\mathcal B$ such that $z\in W\subseteq V_1\cap V_2.$ For the latter, it will suffice (though it isn't necessary) to prove that for any $V_1,V_2\in\mathcal B,$ either $V_1\cap V_2\in\mathcal B$ or $V_1\cap V_2=\emptyset.$ (Do you see why?)
A: Consider the basis for $\mathbb{R}$ as $\mathcal{B}=\{(a,b)\ |\ a,b\in \mathbb{Q}\}$. Show that the usual topology on $\mathbb R$ is same as the topology  generated by $\mathcal{B}$. Since $\mathcal{B}$ is countable so $\mathbb{R}$ is second countable.
