Find a sequence such that for any real number $x$, there exists a subsequence which converges to $x$.
So I pick the sequence of the rational numbers ( we know this is a sequence because the rationals are countable ) so we write
$f : N \to Q$
$n \to (r_1,r_2,r_3, \cdots )$
And then I defined the gubsequence $g : N \to N$ so that $f o g : N \to Q.$
We define g(k) such that $x-1 < r(g(1)) < x+1$
$x-\tfrac12 < r(g(2)) < x+\tfrac12$
And in general,
$x-\tfrac{1}{2^{k-1}} < r(g(k)) < x+ \tfrac{1}{2^{k-1}}$
where $r(g(k))$ is the rational number corresponding to the index $g(k)$ in our list of rationals.
But then, I face a different problem.
How do I ensure that $g(k)$ is a strictly increasing function, i.e , my subsequence doesn't mess up the order the terms appear in?
Note that when you choose the value of $g(k)$, there are only finitely many natural numbers smaller than $g(k)$. But there are infinitely many rationals in $(x - \frac{1}{2^{k}}, x + \frac{1}{2^{k}})$ so there are infinitely many rationals in that interval with an index higher than $g(k)$. Picking the index of such a rational to be $g(k+1)$ allows you to ensure that $g$ is strictly increasing.