A function that's finite everywhere 
Let $c_n \in [0,\infty)$ a sequence of real numbers.If $c_n$ has  $0$ as an accumulation point of it,then prove that there exists an enumeration of $\mathbb{Q}=\{q_1,q_2,,,,q_n,,,,,,\}$ such that the function $f(x)= \sum_{q_n \leqslant x}c_n$ is everywhere finite.

I found this exercise a little bit overwhelming,so here my attemt to prove it:
Let $\epsilon >0$
Then by our hypothesis exists a subsequence $c_{k_n}$ of $c_n$ which converges to $0$.
In other words $\exists n_0 \in \mathbb{N} $ such that $|c_{k_n}|< \epsilon$
Now from definition of limit for $\epsilon =1$ we can find $k_{n_1},,,k_{n_j},,,,$ such that $|c_{k_{n_j}}|<1/2^j$
The set $A=\{k_{n_1},,,k_{n_j},,,,\}$ is infinite and countable,thus \exists a bijective function $f:A \longrightarrow \mathbb{Q}$  
Finally we take the enumeration $\mathbb{Q}=\{f(k_{n_j})\}_{j=1}^\infty$.
Is this proof valid or do i need to come up with a different approach?
 A: Your proof seems perfectly fine. If it were on an exam or something, it might have needed a tad bit more care with some details, but it is well laid out and I suppose it is the straightforward approach.
EDIT (some minor improvements one could make for the sake of pedantry):


*

*When choosing $k_{n_j}$, pick one such that $c_{k_{n_j}}$ is both less than $1/2^j$ and less than any other $c_{k_{n_i}}, i < j$ so that the sequence is strictly decreasing. You should be able to show that such $c_{k_{n_j}}$ always exists.

*It would be a matter of personal taste and a matter of what you talked about in class but if you explicitly created a bijection $\phi $ between say, $\mathbb{N}$ and $\mathbb {Q} $, one could say the bijection used from $A$ to $\mathbb{Q}$ would map $k_{n_j} $ to $\phi(n_j) $ where $\phi $ would have been a bijection $\phi: \mathbb{N} \longrightarrow \mathbb {Q} $ you talked about in class/you created for this exercise

*Lastly you would have to show that the function $f $ defined as in the problem statement (not the one you created) is really finite everywhere for your enumeration (it was the whole point of the exercise). There are several ways to go with it. The first I can think of would be to first show it suffices to prove that $f(q)$ is finite for $q \in \mathbb {Q} $ (why?). Then I would split it into two cases. $q \le 0$ and $q > 0$.

*End your proof with style. It shows confidence I guess. I always write QED or draw that little square $\blacksquare $ or $\square $.
Please note that aside from the third point I was being rather picky, but that looked like what you asked, so I went for it.
