Example 6 Let the function $f_{3}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$
Analysis: First, draw the graph of $f_{0}(x)=|x|$ and translate it downward by 1 unit to get the graph of $y=f_{0}(x)-1$, from which we obtain the graph of $f_{1}(x)=\left|f_{0}(x)-1\right|$ (as shown in Figure 7).

Next, translate the graph of $f_{1}(x)$ downward by 2 units to get the graph of $y=f_{1}(x)-2$, and reflect the part of the graph below the $x$-axis to above the $x$-axis, to obtain the graph of $f_{2}(x)=\mid f_{1}(x)-2 |$ (as shown in Figure 8). The area of the closed figure formed with the $x$-axis is
$$
\begin{array}{l}
\quad S=S_{\text {and }}-S \\
=\frac{6+2}{2} \times 2-\frac{1}{2} \times 2 \times 1 \\
=7 .
\end{array}
$$