【Analysis】If $x_{1}, x_{2}, x_{3}, x_{4}$ have two numbers equal, then $K=0$, which is obviously not the maximum value.
Below, without loss of generality, assume $1 \geqslant x_{1}>x_{2}>x_{3}>x_{4} \geqslant 0$.
Fixing $x_{2}, x_{3}$, to maximize $K$, then $x_{1}=1, x_{4}=0$.
At this point, let $x_{2}=y, x_{3}=z$. Then
$$
K=(y-z)(1-y)(1-z) y z \text {. }
$$

When $y-z$ is fixed, since
$$
\begin{array}{l}
(1-y)+z=1-(y-z), \\
(1-z)+y=1+(y-z),
\end{array}
$$

thus $(1-y)+z$ and $(1-z)+y$ are also fixed.
Furthermore, for two positive numbers with a fixed sum, their product is maximized when they are equal. From $1-y=z$ and $1-z=y$, we can both get $y+z=1$. Therefore, to find the maximum value of $K$, we can set $y=1-z$, combined with $y>z$, we get $z \in\left(0, \frac{1}{2}\right)$. Thus,
$$
K=(1-2 z)(1-z)^{2} z^{2} \text {. }
$$

Let $f(z)=(1-2 z)(1-z)^{2} z^{2}\left(z \in\left(0, \frac{1}{2}\right)\right)$.
Then $f^{\prime}(z)=2 z(1-z)\left(1-5 z+5 z^{2}\right)$.
In the interval $\left(0, \frac{1}{2}\right)$, $f^{\prime}(z)=0$ has a unique root $z_{1}=\frac{5-\sqrt{5}}{10}$. When $z \in\left(0, z_{1}\right)$, $f^{\prime}(z)>0$, $f(z)$ is strictly increasing; when $z \in\left(z_{1}, \frac{1}{2}\right)$, $f^{\prime}(z)<0$, $f(z)$ is strictly decreasing.

Therefore, the maximum value is obtained at $z_{1}=\frac{5-\sqrt{5}}{10}$, which is $f\left(\frac{5-\sqrt{5}}{10}\right)=\frac{\sqrt{5}}{125}$, i.e., $K_{\text {max }}=\frac{\sqrt{5}}{125}$.
Example 3 Real numbers $x_{1}, x_{2}, x_{3}, x_{4} \in [0,1]$. Find $K_{\max }$, where,
$$
\begin{aligned}
K= & \left|x_{1}-x_{2}\right|\left|x_{1}-x_{3}\right|\left|x_{1}-x_{4}\right|\left|x_{2}-x_{3}\right| \cdot \\
& \left|x_{2}-x_{4}\right|\left|x_{3}-x_{4}\right| .
\end{aligned}
$$