Is any infinite set $X$ on the line separable? I am thinking about this question for a while. 
Is any infinite set $X$ on the line separable? 
I think the answer is yes, but to do so, I need to show there exists a countable subset $A$ in $X$ such that $X$ is in the closure of $A$. 
Can anyone outline the proof for me? Thanks.
 A: Yes, assuming by "the line" you mean "the real line $\mathbb{R}$ equipped with the Euclidean topology".
Here is a broader take on this: compare with second countability.  Facts:
1) Every second countable topological space is separable.
2) The converse need not hold, but every metrizable separable space (like $\mathbb{R}$!) is second countable.
3) Every subspace of a second countable space is second countable: just restrict the base.
Thus every subspace of a metrizable separable space is separable.  

Added in response to the OP's comment: Hmm.  In the departments with which I am familiar, students take undergraduate general topology before they take graduate real analysis.  I am not completely confident in the ability of someone to succeed at the latter without having at least some mastery of the former.
Anyway, you asked me if there is a simpler explanation.  Well, "simpler" is subjective, so in my opinion...no, the explanation I gave above is the one which is simplest to me.  But I can explain it in a way which doesn't use second countability: here goes.
For $k \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$, let $I_{n,k} = [\frac{k}{2^n},\frac{k+1}{2^n})$.  
Fix $n \in \mathbb{Z}^+$ and consider the partition $\{ I_{n,k} \}_{k \in \mathbb{Z}}$ of $\mathbb{R}$.  For each $k \in \mathbb{Z}$ such that $I_{n,k} \cap X$ is nonempty, choose one point $a_{n,k}$ of $X$.  Let $A$ be the subset of $X$ consisting of these points $a_{n,k}$.  Then $A$ is countable.  Moreover, for every $x \in X$, we may construct a sequence in $A$ converging to $x$ as follows: let $x_n$ be the unique element $a_{n,k}$ $A$ lying in the same interval $I_{n,k}$ as $x$.  
A: This is just the special case of Pete’s answer that you need for your problem, with a few more details.
There are countably many rationals, so there are countably many open intervals with rational endpoints; list them as $\{I_n:n\in\mathbb{N}\}$. Let $X$ be any subset of $\mathbb{R}$. For each $n\in\mathbb{N}$ there are two possibilities:


*

*$I_n\cap X=\varnothing$: In this case do nothing.

*$I_n\cap X\ne\varnothing$: Let $x_n$ be any point of $I_n\cap X$.


Now let $D=\{x_n:I_n\cap X\ne\varnothing\}$. Clearly $D$ is countable, and I claim that $D$ is a dense subset of $X$, i.e., such that $X=\operatorname{cl}_X D$.
To see this, let $x$ be any point of $X$, and let $(x-\epsilon,x+\epsilon)$ be an open interval centred at $x$. The rationals are dense in $\mathbb{R}$, so there are rationals $p\in(x-\epsilon,x)$ and $q\in(x,x+\epsilon)$. Then $(p,q)$ is an open interval with rational endpoints, so $(p,q)=I_n$ for some $n\in\mathbb{N}$. Moreover, $x\in I_n\cap X$, so $I_n\cap X\ne\varnothing$, and according to $(2)$ above there is a point $x_n\in D\cap I_n$. But $I_n\subseteq(x-\epsilon,x+\epsilon)$, so $x_n\in (x-\epsilon,x+\epsilon)\cap D$. In other words, for each $\epsilon>0$ the $\epsilon$-nbhd of $x$ contains a point of $D$, and therefore $x\in\operatorname{cl}_X D$, as desired.
To connect this with second countability: the family $\{I_n:n\in\mathbb{N}\}$ is a countable base for the Euclidean topology of $\mathbb{R}$. A base for the topology is just a collection $\mathscr{B}$ of open sets such that every open set is a union of some subcollection of $\mathscr{B}$; a space is second countable if its topology has a countable base. If $X$ is any subset of $\mathbb{R}$, $\{X\cap I_n:n\in\mathbb{N}\}$ is a countable base for the subspace topology on $X$, so second countability is hereditary: if a space is second countable, so are all of its subspaces. And the trick that I used above can pretty clearly be applied to any second countable space to get a countable dense subset $-$ which is why every second countable space is separable, and even hereditarily separable.
