Definition of unboundedness in proof that an unbounded monotone increasing sequence is properly divergent. Here is a theorem which can be found in page 92 of Introduction to Real Analysis, Fourth Edition by Robert G. Bartle and Donald R. Sherbert.

If $(x_n)$ is an unbounded increasing sequence, then $\lim(x_n)=+\infty$.
Proof. Suppose that $(x_n)$ is an increasing sequence. We know that if $(x_n)$ is bounded, then it is convergent. If $(x_n)$ is unbounded, then for any $\alpha\in\mathbb R$ there exists $n(\alpha)\in\mathbb N$ such that $\alpha<x_{n(\alpha)}$. But since $(x_n)$ is increasing, we have $\alpha<x_n$ for all $n\geq n(\alpha)$. Since $\alpha$ is arbitrary, it follows that $\lim(x_n)=+\infty$.

Here is my problem:
By definition,
$(x_n)$ is bounded if there exists a real number $M$ such that for all natural numbers $n$, we have $|x_n|\leq M$.
If we negate this we get, $(x_n)$ is unbounded if for every real number $M$, there exists a natural number $n_0$ such that $|x_{n_0}|>M$.
However, while taking into consideration the unboundedness of the sequence, the authors consider $\alpha < x_{n(\alpha)}$ not $\alpha<|x_{n(\alpha)}|$.
Why didn't the authors consider the absolute value of $|x_{n(\alpha)}|$ ?
 A: If an increasing sequence  $(x_n)$ has an upper bound $M$ then $x_1 \leq x_n \leq M$ for all $n$ and this implies that $(x_n)$ is bounded. Since our sequence is not bounded it follows that it has no upper bound. Hence, for any real number $\alpha$ there exits $n(\alpha)$ such that $x_{n(\alpha)} >\alpha$. 
A: For $n$ sufficiently large, $x_n$ must be positive, since the sequence is increasing and unbounded.
Suppose not, then, you'd have $x_1\leq x_n\leq 0$ for all $n\in \mathbb{N}$, since, again, the sequence is increasing. This implies that $|x_n|\leq |x_1|$ for all $n\in \mathbb{N}$ and thus, the sequence would be bounded.
A: 
If $(x_n)$ is unbounded, then for any $\alpha \in \mathbb R$ there exists $n(\alpha)\in\mathbb N$ such that $\alpha < x_{n(\alpha)}$

As you noticed, the authors claim that if the sequence is not bounded, then it is not bounded above. Your irritation is understandable: Of course there are unbounded sequences that are bounded from above (namely the sequences bounded above, but not bounded below).
But since all increasing sequences are bounded below$^1$, an increasing sequence is unbounded if and only if it is not bounded above.

$^1$ If $(x_n)$ is an increasing sequence, then $x_1$ is a lower bound.
