Example 6 Given $A=\{(x, y) \mid | x|+| y | \leqslant 1\}, B=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 1\right\}$, $C=\{(x, y) \mid | x| \leqslant 1, | y | \leqslant 1\}$, then the inclusion relationship between sets $A$, $B$, and $C$ is $\qquad$.
Set $A=\{(x, y)|| x|+| y \mid \leqslant 1\}$ represents the square region in the Cartesian plane enclosed by the lines $x+y= \pm 1$ and $x-y= \pm 1$; set $B=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 1\right\}$ represents the region in the Cartesian plane enclosed by the circle $x^{2}+y^{2}=1$; set $C=\{(x, y)|| x|\leqslant 1| y \mid, \leqslant 1\}$ represents the square region in the Cartesian plane enclosed by the lines $x= \pm 1$ and $y= \pm 1$, all including the boundaries. As shown in the figure, $A \varsubsetneqq B \varsubsetneqq C$.

Key points for solving the problem: The form of the representative elements in a set is an important aspect of understanding the set. For example: set $\{x \mid y= f(x)\}$ represents the domain of the function $y=f(x)$; set $\{y \mid y=f(x)\}$ represents the range of the function $y=f(x)$; and $\{(x, y) \mid y=f(x)\}$ represents the set of points on the graph of the function $y=f(x)$, which is the graph of the function $y=f(x)$.