Upper and lower limit I have the following lemma

Let $\{x_n\}$ be a bounded sequence with lower limit $\underline{x}$ and upper limit $\overline{x}$. For every $\epsilon>0$ there is $N\in\mathbb{N}$ such that for every $n\geq N$ we have
  $$
\underline{x}-\epsilon<x_n<\overline{x}+\epsilon
$$

Why does it follow from this lemma that there is a $K_1\in\mathbb{N}$ such that for all $k\geq K_1$ we have $\overline{x}_k<\epsilon$?
Upper limit $\overline{x}$ is
$$
\overline{x}=\lim_{n\to\infty}\overline{x}_n=\lim_{n\to\infty}\sup\{x_m:m\geq n\}
$$
Lower limit $\underline{x}$ is
$$
\underline{x}=\lim_{n\to\infty}\underline{x}_n=\lim_{n\to\infty}\inf\{x_m:m\geq n\}
$$
Context: I am reading the proof of lemma 3 in this document
and my question is a statement in the proof of lemma 3.
 A: Here's detailed commentary on the proof of Lemma 3. 
Lemma 3 (first half). Let $(x_n)$ be a bounded sequence with upper limit $\bar x$. For every $\epsilon>0$ and every $k\in{\mathbb N}$, there exists $n>K$ such that
$$|x_n-\bar x|<\epsilon.$$ 
Proof: Since the sequence $\bar x_k$ decreases to $\bar x$, there exists $K_1$ such that for all $k\ge K_1$,
$$
\bar x \le \bar x_k<\bar x + \epsilon.\tag{1}
$$
Fix any $k\ge K_1$. Recall that $\bar x_k:=\sup\{x_k, x_{k+1}, x_{k+2},\ldots \}$ by definition. So the left-hand inequality in (1) is a statement that the sup of a certain set exceeds a certain value. Earlier you should have proved a result similar to:

If $\sup A\ge b$ then for every $\epsilon>0$ there is an element $x\in A$ such that $x>b-\epsilon$.

Applying this result to the LHS of (1), we assert that there exists $n\ge k$ such that
$$
x_n >\bar x - \epsilon.\tag2
$$
For this $n$ we also have
$$
x_n\stackrel{(*)}\le \sup\{x_k, x_{k+1}, x_{k+2},\ldots \}=:\bar x_k\stackrel{(1)}<\bar x + \epsilon,\tag3
$$
where (*) follows from the fact that

$x\le \sup A$ for every $x\in A$.

To recap, we've established that for every $k\ge K_1$ there exists $n\ge k$ such that
$$\bar x -\epsilon\stackrel{(2)}< x_n\stackrel{(3)}<\bar x + \epsilon.\tag4$$
Going back to the statement of the Lemma, the claim is that for any $K$ we can find $n$ such that $n$ exceeds $K$ and $n$ satisfies (4). We haven't even talked about $K$ yet, so this constraint on $n$ hasn't been enforced yet, but we can do so by requiring the $k$ we originally fixed to exceed not just $K_1$ but also $K$.
