If $f$ and $g$ are bounded and uniformly continuous prove that $fg$ is uniformly continuous Please check my proof:
Suppose $f(x)$ and $g(x)$ are bounded and uniformly continuous, then we obtain
$|x-y|<\delta \rightarrow |f(x)-f(y)|<\sqrt{M}\sqrt{|x-y|}$
and
$|x-y|<\delta \rightarrow |g(x)-g(y)|<\sqrt{M}\sqrt{|x-y|}$
added those we get
$|x-y|<\delta \rightarrow |f(x)-f(y)||g(x)-g(y)|<\sqrt{m}\sqrt{|x-y|}\sqrt{m}\sqrt{|x-y|}=M|x-y|$
that satisfy Lipschitz condition, therefore $f(x)g(x)$ is uniformly continuous.
 A: If $h=fg$
, you haven't shown that $h$ is Lipschitz, i.e. that there exists $C$ such that $\vert h(x)-h(y) \vert \leq C\vert x-y\vert$ for any $x,y$. Further, you never define $\delta$ or $M$ in your proof. 
We can prove the result using the definition of uniform continuity. Suppose $f,g$ are uniformly continuous on $[a,b]$ and bounded. Let $M$ be the maximum value obtained by either $\vert f \vert$ or $\vert g \vert$. To show that $fg$ is uniformly continuous, let $\varepsilon > 0$. Then there exists $\delta > 0$ such that whenever $\vert x-y\vert < \delta$, we have
$$\vert f(x)-f(y)\vert < \frac{\varepsilon/2}{M+1}$$
$$\vert g(x)-g(y)\vert < \frac{\varepsilon/2}{M+1}$$
Then, if $\vert x-y \vert < \delta$,
$$\vert f(x)g(x)-f(y)g(y)\vert = \vert f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)\vert$$
$$\leq \vert f(x)\vert\vert g(x)-g(y)\vert + \vert g(y)\vert \vert f(x)-f(y)\vert$$
$$< \frac{\varepsilon}{2}\left(\frac{\vert f(x)\vert}{M+1}+\frac{\vert g(x)\vert}{M+1} \right) < \frac{\varepsilon}{2}(1+1)= \varepsilon$$
Thus completing the proof. 
