My personal way to try to show that if $L$ is the set of limits points of $A\subset\Bbb R$ then $L$ is closed Im trying to make the proof on this way: I can define the set of limits points of $A$ as $$L=\{l:(a_n)\to l, a_n\in A\}$$
The hypothesis is that $L$ is closed, i.e. $\forall (l_n)\to m\implies m\in L$. Then I want to show that if 
$$\forall l_n\in L: (l_n)\to m\implies\exists a_n\in A :(a_n)\to m$$
i.e. every limit point of $L$ belongs to $L$ and because is a limit point it must exist some sequence of points of $A$ that converges to it.
Then I started thinking that I can pick a set of $a_n$ from every sequence that converges to some point of the sequence $(l_n)$ close enough to make a convergent sequence to $m$ i.e. taking $|a_k-l_k|<\varepsilon : \lim_{n\to\infty}(a_n)=\lim_{n\to\infty}(l_n)=m$ but because $\varepsilon>0$ the unique way to assure that these quantities design a convergent sequence is making them convergent to zero.
If I choose $\{\varepsilon>0\}:(\varepsilon_k)\to 0$ then I think the proof is done because I will had that 
$$\lim_{n\to\infty}(a_n)=\lim_{n\to\infty}(l_n+\varepsilon_n)=\lim_{n\to\infty}(l_n)+\lim_{n\to\infty}(\varepsilon_n)=m+0=m$$
My question: this is something wrong in this proof or lack of something? Thank you in advance.
 A: You have the right idea. You need to be a little bit careful about what indices you are using. But you can take $\epsilon_n = 1/n$ for example. Then for each $n$ there is some sequence $(a_k)\subset A$, which converges to $l_n$. Hence for each $n$ there is an index $K(n)$ such that $|a_{K(n)}- l_n| < 1/n$. 
Define a sequence $(b_n)\subset A$ by $b_n = a_{K(n)}$. Then given an arbitrary $\epsilon > 0$, we can let $N$ be large enough so that $1/N < \epsilon$; then we have
     $$|b_n - m| \leq |b_n - l_n| + |l_n - m| < \frac{1}{n} + \epsilon < 2\epsilon$$
for all $n > N$. In other words, $b_n \to m$. This is essentially what you wrote down.
A: You seem to have the right idea but it needs to be formalized.
Let $\epsilon > 0$.
By definition of $L$, for all $n$ we can find* $a_n \in A$ such that $|l_n - a_n| < \frac{\epsilon}{2}$.
$(a_n)$ then defines a sequence in $A$. 
Because $\lim(l_n) = m$, there is $N$ such that for $n \geq N$, $|l_n - m| \leq \frac{\epsilon}{2}$.
Then for  $n \geq N$, we get $|a_n - m| = |a_n - l_n + l_n - m|\leq |a_n - l_n| + |l_n - m| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ 
Hence $ \lim(a_n) = m $ so $L$ is closed.
*For each $n$, there is a sequence of elements in $A$ which converges to $l_n$. So there must be an element $a_n$ in this sequence such that $|a_n - l_n| \leq \frac{\epsilon}{2}$
