is there any convergent sub-sequence of a sequence of all rational numbers? Let $(a_n)$ be a sequence of rational numbers, where all rational numbers are terms. (i.e. enumeration of rational numbers)
Then, is there any convergent sub-sequence of  $(a_n)$?
 A: The argument given by Elio Joseph can be refined in order to prove that for any given real number, there always exists a subsequence which converges to it.
For that, let $x \in \mathbb{R}$ be a real number. Consider the interval $[x-1,x+1]$. It contains infinitely many rationals, but in particular it contains one rational, which is some $a_{n_1}$ of our enumeration.
Now, having chosen $a_{n_i}$, we consider the interval $[x-\frac{1}{i+1}, x+\frac{1}{i+1}]$. This contains infinitely many rationals. In particular, we have that it contains one rational which is not one of the first $n_i$ of our enumeration. We take such rational to be our $a_{n_{i+1}}$.
It is clear that $a_{n_i} \to x$.
A: Even better - for every real $x$, there is a subsequence of $(a_n)$ converging to $x$.
Indeed, start by letting $a_{n_1}$ be any element of the sequence in $(x-1,x+1)$. Then, given $a_{n_1},\ldots,a_{n_k}$, since there are infinitely many rationals in each interval, there must exist $n_{k+1}>n_k$ such that $a_{n_{k+1}}\in(x-\frac1{k+1},x+\frac1{k+1})$. Inductively this creates a subsequence $(a_{n_k})$ which converges to $x$.
A: Yes there is. 
Consider the interval $I:=[0,1]$. Since it contains infinitely many rationals, your sequence will have value in $I$ forever.
So you can extract the sub-sequece such that all the values $(a_{\varphi(n)})$ of the new sequence are in $I$ :
$$\forall n \quad a_{\varphi(n)}\in I.$$
Since $I$ is a compact and $(a_{\varphi(n)})$ is a sequence with values in this compact, you can extract of convergent sub-sequence of $(a_{\varphi(n)})$.
Which gives you the convergent sub-sequence of $(a_n)$ you wanted.
