Theorem
Let $I\subseteq \Bbb R$ be a closed, boundet set. Let $A(t)$ be an $n\times n$ continuous matrix on $I$.
Let $X$ be a $n$-vector. Let $F(t,X)=A(t)X$.
Then $F$ is Lipschitz in the variable $X$.
Proof
We have $$||AX-AY||^2=||A(X-Y)||^2=\sum_{i=1}^n\left(\sum_{j=1}^n a_{ij}(x_j-y_j)\right)^2\\ \leq C_n\sum_{i,j=1}^n|a_{ij}|^2(x_j-y_j)^2\leq C_nK^2n||X-Y||^2$$
Where $K>|a_{ij}(t)|\, \forall\, t\in I$. Such $K$ exists by the extreme value theorem.
Let $L^2=C_nK^2n$ and we get that $||F(t,X)-F(t,Y)||=||AX-AY||\leq L||X-Y||$. QED.
I don't get the first and second inequalities: What is $C_n$? What allows you to change the first sum into the one after the inequality? I also don't understand how the author got the second one...
Let's look at $$ \sum_{i=1}^n\left(\sum_{j=1}^n a_{ij}(x_j-y_j)\right)^2 $$ Denote $a_{ij} (x_j - y_j)$ by $u_j$. Denote the vector with coordinates $u_j$ by $u$. Consider a vector with all unit coordinates, denoted by $p$. Then the innermost sum is a dot product $(p \cdot u)$, so it is bounded by product of norms: $$ \left(\sum_{j=1}^n a_{ij}(x_j-y_j)\right)^2 = \left(p \cdot u\right)^2 \le \|p\|^2 \|u\|^2 = \|p\|^2 \left(\sum_{j=1}^n u_j^2\right) = n \sum_{j=1}^n a_{ij}^2 (x_j-y_j)^2 $$ So $C_n$ is simply a square of norm of vector $p$ with unit coordinates, which equals $n$.
Regarding the second inequality, the author has simply bounded all matrix elements $a_{ij}$ by some constant $K$: $$ \sum_{i=1}^n \left( n \sum_{j=1}^n a_{ij}^2 (x_j-y_j)^2 \right) = C_n \sum_{i,j=1}^n |a_{ij}|^2 (x_j-y_j)^2 \leq C_n \sum_{i,j=1}^n K^2 (x_j-y_j)^2 = \\ C_n K^2 \sum_{i=1}^n \sum_{j=1}^n (x_j-y_j)^2 = C_n K^2 n \sum_{j=1}^n (x_j-y_j)^2 = C_n K^2 n \|X-Y\|^2 = n^2 K^2 \|X-Y\|^2 $$