does anyone know the following result? If it holds in this form and any source which presents it? Thanks a lot.
Consider metric space $(X,d_{X})$. Let $f:A \subset (X,d_{X}) \rightarrow \mathbb{R}$ be a Lipschitz continuous function i.e. $\exists K > 0$ such that \begin{align*} |f(x)-f(y)| \leq Kd_{X}(x,y) ~~~~\forall x,y \in A \end{align*} then define functions
$$\bar{f_{1}}(x) := \sup\lbrace f(y) - Kd_{X}(x,y): y\in A \rbrace ~~ \forall x \in X$$ and $$\bar{f_{2}}(x) := \inf\lbrace f(y) + Kd_{X}(x,y): y\in A \rbrace ~~ \forall x \in X$$ it follows that \begin{align*} \bar{f_{1}}, \bar{f_{2}}: X \rightarrow \mathbb{R} \end{align*} are Lipschitz continuous with \begin{align*} |\bar{f_{i}}(x)-\bar{f_{i}}(z)| \leq Kd_{X}(x,z) ~~~\forall x,z \in X \text{ and } \forall i = 1,2 \end{align*} and \begin{align*} \bar{f_{i}}(x) = f(x) ~~~ \forall x \in A \text{ and } \forall i =1,2 \end{align*}
Take $\epsilon>0$ and choose $y_x\in A$ such that $$\overline{f}_1(x)\ge f(y_x)-Kd_X(x,y_x)\ge \overline{f}_1(x)-\epsilon\tag{1}.$$
It follow from $(1)$ and the definition of $\overline{f}_1$ that $$\overline{f}_1(x)-\overline{f}_1(z)\le f(y_x)-Kd_X(x,y_x)+\epsilon-f(y_x)+Kd_X(z,y_x).\tag{2}$$
We conclude from $(2)$ that $$\overline{f}_1(x)-\overline{f}_1(z)\le Kd_X(x,z)+\epsilon.$$
Once $\epsilon$ is arbitrary, we conclude that $\overline{f}_1(x)-\overline{f}_1(z)\le Kd_X(x,z)$. The other inequality is similar, while the fact that $\overline{f}_1$ is a extension of $f$ is trivial. Analogous arguments works also for $\overline{f}_2$. As an reference, take a look in Kirszbraun theorem and in section 2.10.43. of Federer's book.