Example 2 Let $x, y \in \mathbf{R}, M=\max || x+y |$, $|x-y|,|1-x|,|1-y|\}$. Try to find the minimum value of $M$.
Solution: (1) If $xy \geqslant 0$, then
$$
|x-y| \leqslant |x| + |y| = |x + y|.
$$

Thus,
$$
\begin{array}{l}
M = \max \{|x + y|, |x - y|, |1 - x|, |1 - y|\} \\
= \max \{|x + y|, |1 - x|, |1 - y|\}.
\end{array}
$$

Since $M$ is the maximum of $|x + y|, |1 - x|, |1 - y|$, $M$ is not less than their arithmetic mean, i.e.,
$$
\begin{array}{l}
M \geqslant \frac{1}{3}(|x + y| + |1 - x| + |1 - y|) \\
\geqslant \frac{1}{3}|(x + y) + (1 - x) + (1 - y)| = \frac{2}{3}.
\end{array}
$$

Furthermore, when $x = y = \frac{1}{3}$, $M = \frac{2}{3}$.
(2) If $xy < 0$. Then,
$$
\begin{array}{l}
M = \max \{|x + y|, |x - y|, |1 - x|, |1 - y|\} \\
\geqslant \max \{|1 - x|, |1 - y|\} > 1 > \frac{2}{3}.
\end{array}
$$

Therefore, from (1) and (2), we know $M_{\min} = \frac{2}{3}$.
Comment: The main approach is to use
$$
M \geqslant \frac{1}{3}(|x + y| + |1 - x| + |1 - y|),
$$

and then apply the absolute value inequality to transform
$$
|x + y| + |1 - x| + |1 - y|
$$

into a constant.