$f(x)$ non-decreasing then pseudoinverse of $x + f(x)$ is Lipschitz. while studying some proof, I came across the following statement:

Let $f$ be a non-decreasing function defined on closed interval $[a, b]$. Let $\alpha = a + f(a)$ and $\beta=b+f(b)$. We can define $g$ (the "pseudoinverse" of $f$) on $[\alpha, \beta]$ as $g(y)=\sup\left\{x\in[a,b]; x+f(x)\leq y\right\}.$ Then g is non-decreasing and Lipschitz.

The statement is made without any hints. I've tried hard but I can't prove that $g$ is indeed Lipschitz. Honestly, I haven't come up with anything useful...
I'd be really grateful if you could help me.
Thanks a lot!
P.S. I don't know why but I can't write "Hi" at the beggining of my post, so I'm saying hello here.:)
 A: If we let $A_y=\left\{x\in[a,b]:x+f(x)\leq y\right\}$ and $B_y=\left\{x\in[a,b]:y<x+f(x)\right\}$ (notice that $g(y)=\sup A_y$), then $A_y$ and $B_y$ are complementary intervals of $[a,b]$, and $B_y$ is to the right of $A_y$, so $\sup A_y=\inf B_y$, that is,
\begin{align*}
g(y)&=\sup A_y=\inf B_y\tag{$*$}\\
&=\sup\left\{x\in[a,b]:x+f(x)\leq y\right\}=\inf\left\{x\in[a,b]:y<x+f(x)\right\}.
\end{align*}
Now, suppose $y_1\leq y_2$. Then $A_{y_1}\subseteq A_{y_2}$, thus $g(y_1)\leq g(y_2)$, so $g$ is non-decreasing.
Now, let $y_1, y_2$ in $[\alpha,\beta]$. WLOG, we may assume that $y_1\leq y_2$ (otherwise, just reverse the roles) and, since we're analyzing a Lipschitz condition, that $g(y_1)\neq g(y_2)$, that is $g(y_1)<g(y_2)$.
Let $\varepsilon>0$. By $(*)$, we can find $x_1\in B_{y_1}$ and $x_2\in A_{y_2}$ with 
\begin{align*}
x_1-g(y_1)<\varepsilon\qquad\text{ and }\qquad g(y_2)-x_2<\varepsilon.\tag{1}
\end{align*}
Also, since $g(y_1)<g(y_2)$, we can take $x_i$ sufficiently close to $g(y_i)$ so that $x_1<x_2$ and thus $f(x_1)\leq f(x_2)$. Then, since $g$ is non-decreasing and $y_1\leq y_2$,
\begin{align*}
|g(y_2)-g(y_1)|&=g(y_2)-g(y_1)\overset{(1)}{<}2\varepsilon+x_2-x_1=2\varepsilon+x_2+f(x_2)-f(x_2)-x_1\\
&\leq 2\varepsilon+x_2+f(x_2)-f(x_1)-x_1=2\varepsilon+(x_2+f(x_2))-(x_1+f(x_1))\\
&\leq 2\varepsilon +y_2-y_1=|y_2-y_1|+2\varepsilon,
\end{align*}
where the last inequality follows from $x_1\in B_{y_1}$ and $x_2\in A_{y_2}$.
Letting $\varepsilon\to 0$, we obtain $|g(y_2)-g(y_1)|\leq |y_2-y_1|$ for every $y_1,y_2\in[\alpha,\beta]$, so $g$ is Lipschitz.

If we think about smooth, invertible functions, then $\dfrac{df(x)}{dx}\geq 0$, so $\dfrac{d(f(x)+x)}{dx}\geq 1$. Thus, $g$ is the inverse of $f(x)+x$, and it's derivative must satisfy $g'\leq 1$, so that is the reason why we add $x$ to $f(x)$.
A very similar argument would work if we changed $f(x)+x$ by some function $F:[a,b]\to\mathbb{R}$ which is non-decreasing and satisfying $|F(x)-F(y)|\geq c\cdot |x-y|\quad \forall x,y$ for some number $c>0$.
