$\{a_n\}$ is a sequence such that every monotonic subsequence of it converges to the limit $a_{n_k}\to L$. Prove that $a_n\to L$. 
$\{a_n\}$ is a sequence such that every monotonic subsequence of it converges to the limit $a_{n_k}\to L$. Prove that $a_n\to L$.

I was trying to use BW by assuming that this is not the case, but I got stuck.
Any help would be appreciated.
 A: Let us assume that $a_n$ does not converge to $L$. So there exists an $\epsilon_0>0$ with the following property:
$$\forall\ n\in\mathbb{N}\ \exists\ k_nn\geq n:\ |a_{k_n}-l|\geq\epsilon_0$$
So, there exists a subsequence $\left(a_{k_n}\right)$ of $\left(a_n\right)$ such that:
$$|a_{k_n}-l|\geq\epsilon_0$$
But then, $\left(a_{k_n}\right)$ is a sequence itself, so it has a monotonic subsequence, let's say $(a_{k_{m_n}})$. But this is also a subsequence of $(a_n)$, so, it is convergent to $l$, which means that there exists a $n_0\in\mathbb{N}$ such that, for every $n\geq n_0$:
$$|a_{k_{m_n}}-l|<\epsilon_0$$
which is a contradiction to our hypotesis. To explain this a little more, since $(a_{k_{m_n}})$ is a subsequence of $(a_{k_n})$ it follows that:
$$|a_{k_{m_n}}-l|\geq\epsilon_0$$
Now, the contradiction is evident.
So, $a_n\to l$, Q.E.D.
Edit: Let us give a nice proof of the following:

Lemma: If $(a_n)$ is a sequence of real numbers, then it has a monotonic subsequence.

At first, let us give the following:

Definition: If $(a_n)$ is a sequence then we will call a term $a_k$ of the sequence peak point if for every $n\geq k$ we have that:
  $$a_k\geq a_n$$

So, peak points are these terms of the sequence that, when compared to what is after them, are always larger than them.
Now, we can procceed to our proof.
Let us consider the following set:
$$A:=\{k\in\mathbb{N}|a_k\text{ is a peak point of }(a_n)\}$$
Now, consider the case $A$ is an infinite set. Then, it can be written in the form:
$$A=\{a_{k_1},a_{k_2},\dots\}$$
with $k_1<k_2<\dots$, so we can consider the subsequence $(a_{k_n})$ of $(a_n)$. Since every $a_{k_n}$ is a peak point of $(a_n)$, it is evident by the definition we have given, we have that:
$$n<m\Rightarrow a_{k_n}\geq a_{k_m}$$
so, $(a_{k_n})$ is a decreasing subsequence of $(a_n)$ and, hence monotonic.
Consider now the case $A$ is a finite set. Then, it can be written in the form:
$$A=\{a_{k_1},a_{k_2},\dots,a_{k_N}\}$$
for some $N\in\mathbb{N}$. Now, let $m_1=N+1$. Then, there exists a $m_2>m_1=N+1$ such that $a_{m_2}>a_{m_1}$ - if not, then $a_{m_1}=a_{N+1}$ would be a peak point that is larger than $a_N$, which is a contradiction. In the same way, there exists a $m_3>m_2$ such that $a_{m_3}>a_{m_2}$ etc. By induction, we can construct a subsequence $(a_{m_n})$ of $(a_n)$ such that:
$$n_1<n_2\Rightarrow a_{m_{n_1}}<a_{m_{n_2}}$$
or, in other words, $(a_{m_n})$ is increasing, so it is monotonic.
So, in any case, we have found a monotonic subsequence of $(a_n)$ Q.E.D..
It is clear that the abovementioned proof does not make use of anything stronger such as Bolazno-Weiesrtrass etc. It is also interesting that BW can be proved as a corollary of the above Lemma, as follows:

BW: Every bounded sequence has a convergent subsequence.

Indeed, since every sequence (not needed to be bounded; every single one) has a monotonic sequence, let $(a_{k_n})$, then since $(a_n)$ is bounded, the same apllies to $(a_{k_n})$. But a monotonic and bounded sequence is convergent, so $(a_{k_n})$ is convergent Q.E.D..
Hope this cleared it up! :)
