Last inequality for lipschitz - why do we choose this one ?

Question

Let $f:\Bbb R^2 \to \Bbb R \quad f(x,t)=|tx|^{\frac{1}{2}}$

Show that $f$ is lipschitz-continuous on $[a,b] \times [\epsilon, \infty)$ where $\epsilon>0 , a

Here's the solution: \begin{align} |f(t,x)-f(t,y)|& = | |tx|^{\frac{1}{2}} -|ty|^{\frac{1}{2}}| \\ & \le ||tx|^{\frac{1}{2}}| -||ty|^{\frac{1}{2}}| \\ & = |t|^{\frac{1}{2}}|\sqrt x - \sqrt y| \\ & = |t|^{\frac{1}{2}}\frac{|x-y|}{|\sqrt x + \sqrt y|}\\ & \le \underbrace{max\{|a|^{\frac{1}{2}},|b|^{\frac{1}{2}}\}\cdot \frac{1}{2\sqrt \epsilon}}_{=:L} \cdot |x-y| \quad (*)\\ & = L |x-y| \end{align}

Now , not sure how they got the $(*)$ part:

$1)|x-y|$ obviously

$2)x,y<\epsilon \Rightarrow \frac{1}{\sqrt x +\sqrt y} \le \frac{1}{\sqrt \epsilon +\sqrt \epsilon} =\frac{1}{2\sqrt \epsilon} $

$3)$Here I got stuck on: $[a,b] \times [\epsilon, \infty) = \{(t,x)\in \Bbb R^2| a \le t \le b, \epsilon \le x < \infty\} $ and for $a

Maybe someone can explain me the third part

Thanks in advance