Q:

How to sketch $\{(x, y) \in \Bbb R^2 : -1 \leq \text{max}\{|x|,|y|\} \leq 1 \cup 2 \leq \text{max}\{|x|,|y|\} \leq 4\}$

After checking this answer, I did not quite get what I was trying to understand. 
I have a routine exercise which states: Let $f: \Bbb R^2 \rightarrow \Bbb R$, $(x,y) \rightarrow \text{max} \{|x|,|y|\}$ Sketch $f^{-1}([-1,1] \cup [2,4])$
Ok, so here's how the answer goes for this:
By definition, $f ^{-1}([-1, 1] \cup [2, 4]) = \{(x, y) \in \Bbb R^2 : -1 \leq \text{max}\{|x|,|y|\} \leq 1 \text{ or } 2 \leq \text{max}\{|x|,|y|\} \leq 4\}$. Hence, an
appropriate sketch is:

But how do they sketch this?  What are the steps for this? 

A:

Since change of $x$ to $-x$ and $y$ to $-y$ leaves the curves invariant, they symmetric in all quadrant.  So consider only the first quadrant where $x,y \ge 0$.
(1) The first figure is bounded by four lines: $ x=0, y=0, x=1$ and $y=1$. Finally it is
a square with corners $(1,1), (-1,1), (-1,-1),(1,-1)$. So, it is a region with all points inside it and hence it is full grey. 
(2) Similarly, $max{|x|,|y|} \ge 2$ indicates all points outside the square with corners $(2,2),(-2,2),(-2,-2),(2,-2)$ and $max{|x|,|y|}\le 4$ means all points inside the square with corners $(4,4),{-4,4),(-4,-4),(4,-4)$. So the required area is the overlap of these two squares and it is  greyed..
Finally, required region is the union of these two grey regions. 