Originally Posted by
Krizalid yes, it is.
let's break that set into $\displaystyle [0,1]\cup[1,\infty).$
note that the first set is compact, so $\displaystyle f$ is automatically uniformly continuous there, but as for the second set, for each $\displaystyle x,y\ge1$ is $\displaystyle \left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|,$ and this satisfies Lipschitz condition, hence we have uniform continuity on the second set, and finally on the whole set.