Proof verification : Prove set A is a bounded closed set. Let sequence $\langle a_n\rangle$ be a bounded real sequence, and 
$A = \{\alpha \in \mathbb R$ : there exists subsequence of $\langle a_n\rangle$ which converges to $\alpha \}$.
Prove set A is a bounded closed set.
my proof :
As $\langle a_n\rangle$ is bounded, there exists $M \ge 0$ such that $|a_n| \le M$ for all $n \in \mathbb N$. Therefore,  $-M \le a_{n(k)} \le M$ for all $k \in \mathbb N$, and $-M \le \lim_k a_{n(k)} \le M$. Therefore, A is a bounded set.
Now suppose $B = \{a_n : n \in \mathbb N \} $ and let $B^\prime$ be a set of limit point of B. Then, $x \in B^\prime$ is equivalent to the statement that there exists sequence in $B \setminus \{x \}$ which converges to $x$. This means that there exists subsequence of $\langle a_n\rangle$ which converges to $x$. Thus, $x \in A$.
Now suppose $x \in A$. Then there exists subsequence of $\langle a_n\rangle$ which converges to $x$, and there exists a sequence in $ B \setminus \{ x \}$ which converges to $x$. Therefore, $x \in B^\prime$, and $ A = B^\prime$. As $B^\prime$ is a closed set, A is a closed set.
Although I finished my proof, I'm not sure whether my proof doesn't have fallacy. Especially, I get some strange feeling that I made a mistake when I proved if $x \in A$, then $x \in B^\prime$. Did I make mistake? Or do I need more rigorous proof?
 A: To see $A$ is closed it's necessary and sufficient that $A' \subset A$.
So let $x \in A'$. Then fix $m \in \mathbb{N}$. Then there is some $x \neq x_m  \in A$ such that $d(x_m, x) < \frac{1}{2m}$ from $x \in A'$. As $x_n \in A$, there is some subsequence $x_{n_k} \rightarrow x_m$, and applying the definition of convergence ,there is some $M$ such that for all $k \ge M$,  $d(x_{n_k}, x_m) < \frac{1}{2m}$. So some infinitely many terms of the sequence exist such that $d(x, x_n) < \frac{1}{m}$. So 
$$\forall m: \exists_\infty n: d(x_n, x) < \frac{1}{m}$$ 
And this allows us to easily construct a subsequence of $(x_n)$ that converges to $x$, so that $x \in A$ as well. So $A$ is closed.
A: You have an error when you say: If $x\in A$ there exists $(a_n)_n$ in $A$ converging to $x$ and there exists a sequence in $B$ \ $\{x\}$ converging to $x.$ The existence of such a sequence in $B$ \  $\{x\}$  does not follow from what you did previously, and cannot be shown anyway. For example if $a_n=0$ for all $n$ then $A=B=\{0\}.$
Let $(x_n)_n$ be a sequence in $A$ converging to $x.$ Take  $f(1)$ such that $|a_{f(1)}-x_1|<1.$ Take $f(n+1)>f(n)$ such that $|a_{f(n+1)}-x_{n+1}|< 2^{-n}.$  Then $(a_{f(n)})_n$ converges to $x$, so $x\in A.$ 
$f(1)$ exists because $x_1$ is the limit of a subsequence of $(a_n)_n.$
For the existence of $f(n+1)$: Since  $x_{n+1}$ is the limit of a subsequence $(a_{g(m)})_m,$ (with strictly increasing $g$),  take $m'$ large enough that $m''>m'\implies |x_{n+1}-a_{g(m'')}|<2^{-n}.$ And let $f(n+1)=g(m''')$ where $m'''\geq m'$ and $m'''$ is large enough  that $g(m''')>f(n).$
