Prove $f(x) = \sqrt x$ is uniformly continuous on $[0,\infty)$ I have seen a proof in $\sqrt x$ is uniformly continuous
Below shows an alternative proof. Please correct me if im wrong.
Proof:
For any given $\varepsilon >0$,
Let $\delta_1 = \frac{\varepsilon}{2}$, $\forall x,y \in [1,\infty)$ with $|x-y|<\delta_1$
Since $|\sqrt x + \sqrt y|\geq2$
$$|\sqrt x - \sqrt y|  = \frac{|x-y|}{|\sqrt x+\sqrt y|} < |x-y| < \delta_1 = \frac{\varepsilon}{2}$$
Hence, $\sqrt x$ is uniformly continuous on $[1,\infty)$.
$\sqrt x$ is continuous on [0,1] , so $\sqrt x$ is uniformly continuous on [0,1].
So, there exist $\delta_2 > 0$ such that $\forall x,y \in [0,1], |x-y|<\delta_2$, $|\sqrt x -\sqrt y| <\frac{\varepsilon}{2}$
Let $\delta = \min{(\delta_1,\delta_2)}$
$\forall x,y \in [0,\infty)$ with $|x-y|<\delta$,
Case 1: $x,y \in [0,1]$
Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$
Case 2: $x,y \in [1,\infty)$
Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$
Case 3: $x \in [0,1] , y \in [1,\infty]$
$$|\sqrt x-\sqrt y| = |\sqrt x -1+1-\sqrt y| \leq |\sqrt x-1| + |\sqrt y -1| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}= \varepsilon$$
by applying case 1 and case 2.
 A: Let $0 < \epsilon < 1$ and $x,y \in [0,\infty)$ with $|x-y| < \epsilon^2$.
Consider $|\sqrt{x} - \sqrt{y}| = ||\sqrt{x}| - |\sqrt{y}|| \leq |\sqrt{x} + \sqrt{y}|$.
So, $|\sqrt{x} - \sqrt{y}|^2 = |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} + \sqrt{y}| = |x-y|$, that is, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}}$ which implies that $\sqrt{\centerdot}$ is Hölder-continuous.
Finally, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}} < \sqrt{\epsilon^2} = \epsilon$.
Q.E.D.
A: hint
Let $f: \;x\mapsto \sqrt{x} $ and $\epsilon>0$.
$f $ is uniformly continuous at the compact $[0,\frac 1 4] $ since it is continuous.
there exist $\delta_1>0$ such that
$$\forall (x,y)\in [0,\frac 1 4]^2$$
$|x-y|<\delta_1\implies|f (x)-f (y)|<\epsilon $
$f $ is uniformly continuous at $[\frac 1 4,+\infty) $ since it is lipschitzienne.
$(|f (x)-f (y)|=\frac {|x-y|}{f(x)+f(y)|}\leq |x-y|)$.
there exist $\delta_2>0$ such that
$$\forall x,y\geq \frac 1 4$$
Now we treat the case $x\leq \frac 14\leq y.$ 
$f $ is continuous at $\frac 14$, thus
there exist $\delta_3>0$ such that
$$|x-\frac 1 4 |<\delta_3\implies |f (x)-\frac 1 2|<\frac {\epsilon}{2} $$.
so if $x\in [0,\frac 1 4] $ and $y\in [\frac 1 4,+\infty) $ with $|x-y|<\delta_3$ then
$|x-\frac 1 4|<\delta_3$ and $|y-\frac 1 4|<\delta_3$
thus
$$|f (x)-\frac 1 2|<\frac {\epsilon}{2} $$ and
$$|f (y)-\frac 1 2|<\frac {\epsilon}{2} $$
$$\implies $$
$$|f (x)-f (y)|\leq|f (x)-\frac 1 2|+|f (y)-\frac 1 2|<\epsilon $$
Now we take $$\delta=\min (\delta_1,\delta_2,\delta_3) $$
Can you finish.
