Given $a,b$ $\in$ $\mathbb{R}$ ($a \lt b$) and functions $y,z \in C\big([a,b],\mathbb{R}\big)$. Prove if $f \in C^{1}\big(\mathbb{R}^{2},\mathbb{R}\big)$, then $\exists K \gt 0$ as:
$$ |f(s,y(s)) - f(s,z(s))| \le K|y(s) - z(s)| \quad \forall s \in [a,b] $$
Note that $c(s)=\min\{y(s), z(s)\}$ and $d(s)=\max\{ y(s), z(s)\}$ are continuous functions of $s$. Fix $I_s=[c(s),d(s)]$ and $I=\bigcup_{s\in [a,b]}I_s$. If $c=\inf_{s\in [a,b]}c(s)$ and $d=\sup_{s\in [a,b]}d(s)$ then $I\subset [c,d]$.
It suffices that you prove that the function $\mathbb{R}^2\ni (s,y)\mapsto f(s,x)\in \mathbb{R}$ is uniformly lipschitz in the second variable $x$ in the set $ [a,b]\times I $. Here the constant $K$ will depend only on the functions $y(s)$ and $z(s)$ but not $s\in[a,b]$. That is, there is a constant $K>0$ that does not depend on $s\in[a,b]$ such that $$ |f(s,u) - f(s,v)| \le K|u - v|. $$ Recall that a function $(s,x)\mapsto f(s,x) $ is a uniformly lipschitz in the second variable if, only if, $$ K=\sup_{s}\sup_{u,v} \frac{|f(s,u)-f(s,v)|}{|u-v|}<\infty. $$ By mean value theorem there is $\tau_s\in [0,1]$, whit $w_s= \tau_sc_s+(1-\tau_s)d_s\in [c_s,d_s]$ and $s=\tau_ss+(1-\tau_s)s$, such that \begin{align} \sup_{u,v\in [c_s,b_s]} \frac{|f(s,u)-f(s,v)|}{|y(s)-z(s)|} =& \sup_{u,v\in [c_s,b_s]} \frac{\big| D_1 f(s,w_s)\cdot [s-s)]+ D_2 f(s,w_s)\cdot [d_s-c_s]\,\big|}{|y(s)-z(s)|} \\ =& \sup_{u,v\in [c_s,b_s]} \frac{\big|D_2 f(s,w_s)\cdot [d_s-c_s]\,\big|}{|y(s)-z(s)|} \\ =& \sup_{u,v\in [c_s,b_s]} \big|D_2 f(s,w_s)\big| \\ \leq & \sup_{u,v\in [c_s,b_s]} \sup_{w\in [c_s,d_s]}\big|D_2 f(s,w)\big| \\ = & \sup_{w\in [c_s,d_s]}\big|D_2 f(s,w)\big| \\ \leq & \sup_{w\in [c,d]}\big|D_2 f(s,w)\big| \end{align} for $[c,d]=\bigcup_{s\in [a,b]}[a_s,b_s]$. Since $[a,b]\ni s \mapsto \sup_{w\in [c,d]}\big|D_2 f(s,w)\big| $ is a continuous function then $$ \sup_{s\in [a,b]}\sup_{w\in [c,d]}\big|D_2 f(s,w)\big|<\infty. $$