Question: Let $\{x\_n\}\_{n\in \mathbb{N}}\subset \mathbb{R}$ be a bounded sequence. We define $\forall N \in \mathbb{N}$:
$$a\_N := inf\{x\_n \mid n \geq N\}$$
$$b\_N := sup\{x\_n \mid n \geq N\}$$
$$l := \lim\_{N \to \infty} a\_N$$
$$L := \lim\_{N \to \infty} b\_N$$
1. Prove that $\{x\_n\}\_{n\in \mathbb{N}}$ converges $\iff$ $l = L$.  

2. Prove that there exist subsequences such that one converges to $l$ and the other to $L$.

I've proved $a\_N$ is monotonic increasing, $b\_N$ is monotonic decreasing, that $a\_N \leq x\_N \leq b\_N$, $\forall N\in \mathbb{N}$, and that $l\leq L$. It seems that the right-to-left impliation of question 1 can be proved using the sandwich rule...

I'd like a full answer to question 1 and hint(s) for question 2.

Answer: 
1) $\Longrightarrow$ 

Let $\left(x\_{n}\right)\_{n}$ converge to
$x$. 

For $\varepsilon>0$ some $n\_{\varepsilon}$ exists such that
$n\geq n\_{\varepsilon}\Rightarrow\left|x-x\_{n}\right|<\varepsilon$.

Then $a\_{N}\geq x-\varepsilon$ and $b\_{N}\leq x+\varepsilon$ for
each $N\geq n\_{\varepsilon}$ and consequently $x-\varepsilon\leq l\leq L\leq x+\varepsilon$.

This is true for any $\varepsilon>0$, hence $l=x=L$. 

1) $\Longleftarrow$ 

The 'sandwichrule' is fine here: $a\_{n}\leq x\_{n}\leq b\_{n}$ with $a\_{n}\uparrow l$, $b\_{n}\downarrow L$ and $l=L$.

2)

Note for any positive integer $k$ the set $I\_{k}:=\left\{ n\in\mathbb{N}\mid l\leq x\_{n}\leq l+\frac{1}{k}\right\} $
is infinite. 

This makes it possible to choose indices $n\_{k}$ such
that $n\_{k}\in I\_{k}$ and $n\_{1}<n\_{2}<\cdots$. 

Then subsequencence
$x\_{n\_{k}}$ converges to $l$. 

For $L$ their is a similar proof.
2) consider the sequence $a\_N+1/N$ resp. $b\_N-1/N$ and their relations to the partial sequence $x\_N,x\_{n+1},...$.

The second part of 1) is actually a consequence of 2), since all subsequences of a convergent sequence have the same limit.