If $\{s_n\}$ is a sequence containing all rationals, then every real number is a subsequential limit? This is Example 3.18 of Rudin's Principles of Mathematical Analysis.  I'm not sure if I understand it correctly.  In particular, I wondered why the statement (as in the title) didn't have to specify how the rationals were ordered in the sequence $\{s_n\}$.  I listed below my attempt to validate the statement in the title with an example.  I'd appreciate it if someone can confirm it or correct me.
For simplicity (WLOG), pick the real number '$2$'.  We construct a subsequence $\{s_{n_k}\}$ converging to '$2$' as follows:


*

*Pick $s_{n_1}=1.$  

*Pick $s_{n_2}\in [2-\frac{1}{2}, 2]$, with $n_2>n_1$.  (This is feasible because there are infinitely many rationals in $[2-\frac{1}{2}, 2]$.)  

*Similarly, pick $s_{n_3}\in[2-\frac1{3},2].$  (Note that $s_{n_3}$ need not be greater than $s_{n_2}$.)

*Continue this process ($s_{n_k}\in [2-\frac1{k},2]$) to construct the rest of $\{s_{n_k}\}$ for $k=4, 5, 6,...$


Is this a valid example to show that the order of the rationals in $\{s_n\}$ doesn't matter and that every real number must be a subsequential limit of $\{s_n\}$?  I'd appreciate confirmation, corrections or comments.  Thanks a lot!

You have the right idea in saying that yours intervals contain an infinite number of rationals. Indeed, if you seek a subsequence of rationals $s_{n_k}$ converging to a real number $x$, it could happen that some elements of your sequence appearing in the beginning of the sequence are near to $x$ (or even $x$ itself if it is a rational). But that does not matter: it remains an infinity of rational near to $x$, for you only have removed in the first terms a finite number of rationals.
So your formalization is correct, and the key fact is indeed the infiniteness of rationals contained in each interval $[x-\varepsilon, x+\varepsilon]$.
