Show that $f$ is continuous mathematically. 
Let $f:[0,\infty)\to \mathbb{R}$ be given by $f(x)=\sqrt{x}$. Show
  that it is continuous.

This is taken from Example 3.7 on <link> page 22 on the paper. It has shown that it is continuous at $c=0$ and $c>0$ seperately, which I understood why. I wondered how I would prove it if I got a similar problem from my teacher without showing how I have processed it (that is, how the values of $\delta>0$ were found). I thought about the following way,

Let $\epsilon>0$ be given and let $c\in [0,\infty)$. Choose
  $$\delta=\begin{cases} \epsilon^{2} & \text{ if } c=0 \\ 
 \epsilon\sqrt{c} & \text{ if } c>0. \end{cases}$$ Assume $c=0$. If
   $\left | x-c \right |=\left | x-0 \right |<\delta$, then $$\left |
 f(x)-f(c) \right |=\left | \sqrt{x}-0 \right |=\left | x-0 \right
 |^{1/2}<\delta^{1/2}=\epsilon.$$ Assume $c>0$. If $\left | x-c \right
 |<\delta$, then $$\left | f(x)-f(c) \right |= \left |
 \sqrt{x}-\sqrt{c} \right |=\left | \frac{x-c}{\sqrt{x}+\sqrt{c}}
 \right |\leq \frac{1}{\sqrt{c}}\left | x-c \right |
 <\frac{1}{\sqrt{c}}\delta=\epsilon.$$ This proves that $f$ is
  continuous.

Is this correct, or is there a better way to show it mathematically?

As to whether there is a "better" way, it actually seems to me that proving continuity via $\epsilon$-$\delta$ specifically for the square root looks a bit awkward.
There is the elegant argument that the inverse of a bijective continuous function $f$: $[a, b]\longrightarrow f([a, b])$ is continuous, and the general proof sort of writes itself.
Then, applying this to $x\mapsto x^2$ on $[0, r]$, $r>0$, gives the result immediately.
