Prove theorem of Bolzano-Weierstrass by least upper bound property Let be $(a_n)_{n\in\mathbb{N}}$ a bounded sequence of real numbers and we want to show that it contains a convergent subsequence.

My approach:
We know, by the least upper bound property, that the set $M:=\{a_n\mid n\in\mathbb{N}\}$ has a supremum $S$. As $S$ is the least upper bound of $M$, for each $\epsilon>0$ there exists a $a_n$ such that $S-\epsilon<a_n<S+\epsilon$.
Let be $\epsilon>0$ then there exists a $k_0\in\mathbb{N}$ such that $\frac{1}{k_0}<\epsilon$. We define a subsequence $\left( a_{n_k}\right)_{k\in\mathbb{N}}$ as follows:
There exists a $a_n$ which we set $a_{n_{k_0}}:=a_n$ such that
$$
S-\frac{1}{k_0}<a_{n_{k_0}}<S+\frac{1}{k_0}.
$$
For all $k>k_0$ it follows that
$$
S-\frac{1}{k_0}<S-\frac{1}{k}<a_{n_k}<S+\frac{1}{k}<S+\frac{1}{k_0}\\
\implies |a_{n_k}-S|<\frac{1}{k}<\epsilon.
$$
Hence, $(a_n)_{n\in\mathbb{N}}$ is convergent.

My tutor said that it is flawed!? I have spent hours revising it and couldn't find any mistakes?
 A: The flaw is that the selected term may always be the same one. In particular if the difference of one term with the other ones is greater than a fix number.
Example. Take $\{a_n\}$ the always vanishing sequence except that $a_1 =1$. Then $S = 1$ but you won't be able to select a converging subsequence to $1$. Whatever $k \in \mathbb N$ you take, the only $a_n$ satisfying $
\vert a_n - 1 \vert \lt 1/k$ is $a_1$. Remember than for a subsequence $\{a_{n_k}\}$, $n_k$ has to be strictly increasing with $k$.
You can however build a subsequence converging to $\limsup\limits_{n} a_n$.
A: Other comments and answers have told you where your errors are.
The existential problem with this question is the following: When you have seen just a finite number of the $a_n$'s you have no idea where a point of convergence could be. This means that you have to encompass the full sequence $(a_n)_{n\geq1}$ at each step of the process, in order to find a point of convergence.
We may assume $0\leq a_n\leq1$ for all $n$. Let $S$ be the set of all $x\in{\mathbb R}$ such that for infinitely many $n$ we have $a_n\geq x$. Then $0\in S$, and $S$ is bounded above by $1$. The set $S$ therefore has a supremum $\sigma\in{\mathbb R}$. I claim that there is a subsequence $k\mapsto a_{n_k}$ with $\lim_{k\to\infty} a_{n_k}=\sigma$.
Proof. Put $n_1:=1$, and assume that for a $k\geq1$ the "selection" $n_k$ has been constructed such that $|a_{n_k}-\sigma|\leq{1\over k}$. We then  shall find an $n_{k+1}>n_k$ such that $$|a_{n_{k+1}}-\sigma|\leq{1\over k+1}\ .\tag{1}$$
By definition of the sup there is an $s\in S$ with $s>\sigma-{1\over k+1}$, and there are infinitely many $a_n\geq s$. On the other hand there are only finitely many $a_n>\sigma+{1\over k+1}$. It follows that there is an $n_{k+1}>n_k$ satisfying $(1)$.
