Here's my attempt: Let $\varepsilon > 0$. Further assume $x>y$. Thus, $| f(x) - f(y) | = \sqrt x - \ \sqrt y < \sqrt{x-y} < \varepsilon$ iff $x-y < \varepsilon ^2$. Hence, we let $\delta = \varepsilon ^2$ and thus if $|x-y |< \delta$, we have $| f(x) - f(y) | < \varepsilon $.
Suppose that the function is Lipschitz. Then there is $M>0$ such that for all $x,y \in [0, \infty)$, we have $|f(x) - f(y)| \le M |x-y|$. Now, choose $a \in \mathbb{R}^{+}$ such that $a<\frac{1}{M^2}$. Now, let $x=a$ and $y=0$, then we have $|f(a)-f(0)| \le M |a|$ iff $\frac{1}{M^2} \le a$. A contradiction!
Is my proof correct?
It's good, and well-presented.
My only criticism is the following line of working:
My mind flagged it as something not obviously true, and so I had to verify it for myself. While it is true (under the given assumption of $x > y$), it's still a good sign that it could use a little more justification.