Proving that copulas are Lipschitz continuous A copula is a function $C:[0,1]^2\to[0,1]$ such that $C(x,0)=C(0,x)=0$ for all $x\in[0,1]$, $C(x,1)=C(1,x)=x$ for all $x\in[0,1]$, and
\begin{equation}\label{ineq}
 C(x_2,y_2)-C(x_1,y_2)-C(x_2,y_1)+C(x_1,y_1)\ge0\tag{*}
\end{equation}
for all $(x_1,y_1), (x_2,y_2)\in[0,1]^2$ with $x_1\le x_2$ and $y_1\le y_2$. I am trying to show that a copula is Lipschitz continuous in the following sense:
$$
 |C(x_2,y_2)-C(x_1,y_1)|
 \le|x_2-x_1|+|y_2-y_1|
$$
for all $(x_1,y_1),(x_2,y_2)\in[0,1]^2$.
Suppose that $x_2\ge x_1$ and $y_2\ge y_1$. Using the definition of a copula, we have that
$$
x_2-x_1-C(x_2,y_1)+C(x_1,y_1)\ge0
$$
and
$$
y_2-y_1-C(x_2,y_2)+C(x_2,y_1)\ge0.
$$
Adding these two inequalities, we obtain
$$
C(x_2,y_2)-C(x_1,y_1)
\le x_2-x_1+y_2-y_1.
$$
Since copulas are increasing in each argument, $
C(x_2,y_2)
\ge C(x_2,y_1)
\ge C(x_1,y_1)
$
so that
$
C(x_2,y_2)-C(x_1,y_1)\ge0
$
and hence
$$
|C(x_2,y_2)-C(x_1,y_1)|\le x_2-x_1+y_2-y_1,
$$
when $x_1\le x_2$ and $y_1\le y_2$.

We also need to consider the case when $x_1\le x_2$ but $y_1\ge y_2$. If I understand correctly, inequality \eqref{ineq} is not valid in this case and I am not sure how to proceed. How can we proceed with the proof when $x_1\le x_2$ but $y_1\ge y_2$?

Any help is much appreciated!

We have $x_1\le x_2$ but $y_1\ge y_2$. Then either  $C(x_2, y_2)\geq C(x_1, y_1)$ or $C(x_2, y_2)\leq C(x_1, y_1)$. Take the case when $C(x_2, y_2)\geq C(x_1, y_1)$.  Then, you want to prove that
$$
C(x_2, y_2)- C(x_1, y_1)\leq x_2-x_1 + y_1-y_2.
$$
From the marginal monotonicity you get
$$
C(x_2, y_2)- C(x_1, y_1)\leq C(x_2, y_2)- C(x_1, y_2). 
$$
Now, you are in the previous setting, i.e. holds that
$$
C(x_2, y_2)- C(x_1, y_2)\leq x_2-x_1. 
$$
Hence, $C(x_2, y_2)- C(x_1, y_1)\leq x_2-x_1. $
In the other case when $C(x_2, y_2)\leq C(x_1, y_1)$, we obtain the same
$$
C(x_2, y_2)- C(x_1, y_1)\leq y_1-y_2. 
$$
Hence, together,
$$
C(x_2, y_2)- C(x_1, y_1) \leq \max(x_2-x_1,y_1-y_2) \leq x_2-x_1+y_1-y_2. 
$$
