Show f(x):=sqrt(x) is uniformly continuous on [0,1]

Question

Show $f(x):=\sqrt{x}$ is uniformly continuous on $[0,1]$.

What I did: Need to show that $\forall \varepsilon>0: \exists\delta>0$ such that
$\forall x,y\in(0,1): |x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon$

Choose $\delta=\varepsilon^2$.

then for $x,y\in(0,1)$ with $|x - y| < \delta$,

$|f(x)-f(y)|=|\sqrt{x}-\sqrt{y}|$

$|\sqrt{x}-\sqrt{y}|\le|\sqrt{x}+\sqrt{y}|$

$|\sqrt{x}-\sqrt{y}|^2\le|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}| =|x-y|< \varepsilon^2$

$|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}<\sqrt{\varepsilon^2}=\varepsilon$

Answer 1

Your solution is correct. I would probably not give you full marks for it for not explaining why the inequalities you use are true.

Also, you were required to prove the uniform continuity on $[0,1]$, but only used $(0,1)$ in your argument (it still works for $x=0$ or $y=0$, but you didn't say so).