Let $f: \Bbb R \to [0,\infty)$ be a continuous function such that $g(x)=(f(x))^2$ is uniformly continuous. Which of the following is always true? 
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?
EDIT:
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 $f$?
 A: Hint. You want to prove that $f(x) = \sqrt{g(x)}$ is uniformly continuous. The statement of this looks like "$|f(x) - f(y)| \leq \epsilon$ whenever...(condition on $x$ and $y$ being close enough)." Since $g$ is known to be uniformly continuous, you can always write "$|g(x) - g(y)| \leq \alpha$ whenever..."
So it will be enough for you to find some function $h \colon (0,+\infty) \to (0,+\infty)$ with the property that whenever $|g(x) - g(y)| \leq \alpha = h(\epsilon)$, you necessarily have $|f(x) - f(y)|\leq \epsilon$. Since the only thing you really know is that $f(x)$ is the square root of $g(x)$, this clarifies what conditions you would want the function $h$ to satisfy. Now your job is to find a function $h$ that works.
Essentially, the proof amounts to showing that the square root function is uniformly continuous.
Edit. Do you have a theorem that says a composite of uniformly continuous functions is uniformly continuous? If so, all you need to do is prove that $j(t) = \sqrt{t}$ is uniformly continuous, since $f(x) = j(g(x))$. (If not, you will need to include a small additional step that amounts to proving this is the special case you're dealing with.)
To prove the uniform continuity of $j$, it will be enough to prove that, for $a, b \geq 0$,
$$|a - b| \leq \epsilon^2 \Longrightarrow |\sqrt{a} - \sqrt{b}| \leq \epsilon.$$
An alternative is to show that $j$ has a Lipschitz constant on $[1,+\infty)$ and to note that $[0,1]$ is a closed bounded interval, hence $j$ must be uniformly continuous there.
A: in your second example (where $f(x)=1$),
$f$ is a continuous function and $g(x)= 1$ is uniform continuous. But note that, $f(x)$ is also uniform continuous ! Hence option b) is also discarded! Therefore, option c) MUST be right :p 
