A. $f$ is bounded
B. $f$ may not be uniformly continuous
C. $f$ is uniformly continuous
D. $f$ is unbounded.
Let $f(x)=\sqrt x$. Then $g(x)=x$ is uniformly continuous and unbounded. Hence option A can not be true.
Let $f(x)=1$. Then $g(x)=1$ is uniformly continuous and bounded. Hence the option D can not be true.
How should I check uniform continuity?
Since $g$ is uniformly continuous, hence for a given $\epsilon \gt 0$, $\exists \delta \gt 0$ depending only upon $\epsilon$ such that $|g(x)-g(y)| \lt \epsilon$ whenever $|x-y| \lt \delta$ $\forall x,y$.
$\Rightarrow |\sqrt g(x) -\sqrt g(y)||\sqrt g(x)+\sqrt g(y)|\lt \epsilon$ whenever $|x-y| \lt \delta$.
$\Rightarrow |f(x)-f(y)||f(x)+f(y)|\lt \epsilon$ whenever $|x-y| \lt \delta$.
What should be my next step now? How can I use the continuity of $f$?