Proof Verification: Prove $\sqrt{x}$ is uniformly continuous on [0, $\infty$) Proof:
Fix $\epsilon \gt 0$. We want to find $\delta \gt 0$ such that:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}| \lt \epsilon$$
$$\phantom{2000i11111}\Rightarrow |\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \epsilon$$
If $|x-a|\lt \delta$ then $|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}$ for $\delta \lt 1$
Thus, choose $\delta = min\{1, (\sqrt{a+1}+\sqrt{a})\epsilon\}$. Then we have:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}|=|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}=\frac{(\sqrt{a+1}+\sqrt{a})\epsilon}{\sqrt{a+1}+\sqrt{a}}=\epsilon$$
 A: Simply, if $|x - a| < \delta(\epsilon) = \epsilon^2$, then, since $|\sqrt{x} - \sqrt{a}| \leqslant |\sqrt{x} + \sqrt{a}|$, we have  
$$|\sqrt{x} - \sqrt{a}|^2 = |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}-\sqrt{a}|\leqslant |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}+\sqrt{a}| = |x - a|< \epsilon^2\\ \implies|\sqrt{x} - \sqrt{a}|< \epsilon$$
A: The $\delta$ depends on $a$, so the reasoning does not go through.
One can do it in this way. Given $\epsilon>0$, we know that $\sqrt{x}$ is continuous on $[0,1]$, and $[0,1]$ is compact, so $\sqrt{x}$ is uniformly continuous on $[0,1]$, so there exists a $\delta_{1}>0$ such that $|x-y|<\delta_{1}$, $x,y\in[0,1]$, then $|\sqrt{x}-\sqrt{y}|<\epsilon$.
$\sqrt{x}$ is continuous at $x=1$, there exists some $\delta_{2}>0$ such that $|x-1|<\delta_{2}$, then $|\sqrt{x}-1|<\epsilon$.
Now take $\delta=\min\{\delta_{1},\delta_{2},\epsilon/2\}$, when $x,y\in[0,\infty)$ with $|x-y|<\delta$, then we have several cases.
Case I. Both $x,y\in[1,\infty)$, then $|\sqrt{x}-\sqrt{y}|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}\leq|x-y|<\epsilon$.
Case II. $x\in[1,\infty)$, $y\in[0,1]$, then $|y-1|=1-y=1-x+x-y\leq x-y<\delta_{2}$, and also that $|x-1|=x-1=x-y+y-1\leq x-y<\delta_{2}$, so $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}-1|+|\sqrt{y}-1|<\epsilon/2+\epsilon/2=\epsilon$.
Case III. Both $x,y\in[0,1]$, then $|x-y|<\delta_{1}$, so $|\sqrt{x}-\sqrt{y}|<\epsilon$.
