product of two uniformly continuous functions is uniformly continuous

There is a nice way:

Hint: Try to show that if f, g are Uniformly continuous, so are $f \pm g$ and $f^2$. Then observe that $fg = 0.5((f+g)^2 - f^2 -g^2)$. Hope this helps.

Let $f$ and $g$ be bounded functions. Hence there are $c,d\in\mathbf{R}$ such that $c,d>0$, $\vert f(x) \vert < c$ and $\vert g(y) \vert < d$ for every $x,y\in(a,b)$. Let $\epsilon>0$.

Since $f$ is uniformly continuous on $(a,b)$, $\exists\delta_f(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_f$, $|f(x)-f(y)|<\epsilon/2d$.

Since $g$ is uniformly continuous on $(a,b)$, $\exists\delta_g(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_g$, $|g(x)-g(y)|<\epsilon/2c$.

Let $\delta = min\{\delta_f,\delta_g\}$. Hence, for all $x,y\in(a,b), \vert x-y \vert<\delta \Rightarrow |g(x)-g(y)|<\frac{\epsilon}{2c}$ and $|f(x)-f(y)|<\frac{\epsilon}{2d}$. Since $\vert f(x) \vert < c$ and $\vert g(y) \vert < d$ for every $x,y\in(a,b)$, it also implies that

$$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \leq |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)| < c.\frac{\epsilon}{2c} + d.\frac{\epsilon}{2d} = \epsilon$$

Finally we have $|f(x)g(x)-f(y)g(y)| < \epsilon$ and $f.g$ is uniformly continuous if $f$ and $g$ are bounded functions.

$f$ and $g$ are continuous on $[a,b]$ hence bounded

try to show :$\lim_{x\rightarrow a+} f(x)$ and $\lim_{x\rightarrow b-} f(x) $ exist as finite limits.