Inequalities involving absolute value 
Let $x,y\in \Bbb R$.\begin{align}
|x|-1\le|y| & \iff|x|-1\le y  \text{ or }|x|-1\le -y \\
 & \iff |x|-1\le y \text{ or } -|x|+1\ge y \\ 
 & \implies |x|-1\le 1-|x| \\
 & \implies |x|\le 1 \\ 
 \end{align}

Is the above calculation correct? I have doubt about $|x|-1\le 1-|x|$ step, I mean, how does 'or' lead us there?
Also, how do we decide if 'and' is appropriate, or, 'or': $|x|\le2\iff  x\le2 \text{ and } -x\le2$ while, $|x|\ge2\iff x\ge2\text{ or }-x\ge2$. Why does one have and while other have or?
 A: The absolute value function can look odd at first, but you can get the hang of it pretty quickly.
First off, I'd like to point out the correct approach for the second part. Think carefully about what the following means:
$$ |x|\le 2$$
You can read it aloud, too. It says that the number $x$, taken with no sign, is smaller than $2$. You can clearly see that this is equivalent to:
$$ - 2 \le x \le 2 $$
This means $x$ must be greater than minus two and lesser than two. If you considered all numbers greater than minus two or lesser than two, any number would go (since any number is either greater than minus two or lesser than two, or in between).
Now look at the other condition.
$$ |x|\ge 2$$
It says that the number $x$, taken with no sign, is bigger than $2$. You can clearly see that this is equivalent to:
$$ x \le - 2 \quad \text{or} \quad x \ge 2 $$
There is no way other than using or here. Why is the or necessary here? Well, think about any number that is simultaneously less than minus two and bigger than two...
About your inequality, think about it as you would with any function. First, express $y$ as a function of $|x|$, and then go straight to listing all possible cases as $x$ varies. Can you take it from here? 
