Is it valid to take $|- \infty | = \infty$? Is it valid to take $|- \infty | = \infty$?
or is the absolute value e.g. not defined for infinity?
Particularly,
if one wishes to argue that operator $f(x)=x$ is not bounded below on $\mathbb{R}_{-}$, then the definition for bounded belowness for operators says must be $\beta > 0$ s.t.
$$\| T x \| \geq \beta\|x\|$$
(and here the norm is abs)
but then if $x \rightarrow -\infty$?
 A: It is indeed true that $\lim_{x \to -\infty} |x| = \infty$ and can be elementary checked.
As for the operator note, it is indeed true, since either $-\infty$ or $\infty$ give an expression of the type $\infty \geq \beta \infty$ with $\beta >0$, thus the operator is unbounded.
Be careful when referring to boundedness of operators :
Let $T:X \to Y$ be a linear operator. A bounded linear operator is generally not a bounded function. Trully, in many cases one can find a sequence $x_{k} \in X$ such that $ \|Tx_{k}\|_{Y}\rightarrow \infty $. Instead, all that is required for the operator to be bounded is that :
$$\frac{\|Tx\|}{\|x\|} \leq M < \infty$$
Thus, the expression $\|Tx\| \geq \beta \|x\|$ does not mean bounded belowness as we would say when talking about a common function. 
A: Yes, that's correct. Note that in your example you may take $\beta=1$ since $\infty\ge 1\infty$.
A: Beware that in $n$ dimensions (as the use of the norm seems to indicate), $x\to\infty$ and $x\to-\infty$ are meanigless/useless, because vectors can escape to infinity in many different ways. Such as $(t,-t,3,-\sin t,-t)$ where $t$ tends to infinity.
So all you consider is the norm, a scalar, that is always positive.
