real analysis - Show that $f : [0, +infty) rightarrow mathbb{R}$, with $f(x) = sqrt{x}$, is uniformly continuous

Let $f : [a, +\infty) \rightarrow \mathbb{R}$. Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is uniformly continuous, and $f: [b, +\infty) \rightarrow \mathbb{R}$ is uniformly continuous.
Show that $f$ is uniformly continuous.
Then, use this idea to show that $f : [0, +\infty) \rightarrow \mathbb{R}$, with $f(x) = \sqrt{x}$, is uniformly continuous

I understand that we can't use a theorem to show continuity on the whole domain, so instead we cut the domain into smaller portions to apply certain theorems. In the second part I can see why we need to cut up the interval into one which is closed and one which is bounded.