So I have sketched some proof and would love some feedback:
We'll divide the proof into two parts:
First assume $A = (a,b)$ for some $a,b \in \mathbf R$.
$f,g$ are uniformly continuous in $A$ $\implies$
$\exists \lim_{x \to a^+} f(x), \lim_{x \to a^+} g(x), \lim_{x \to b^-} f(x), \lim_{x \to b^-} g(x)$
$\implies \exists \lim_{x \to a^+} fg(x), \lim_{x \to b^-} fg(x)$.
In addition, in the interval $(a,b)$: $f,g$ uniformly continuous $\implies$ $f,g$ are continuous.
Therefore, from arthimetics of continuous functions we have $fg$ is continuous in $[a,b]$.
From Cantor's theorem we have $fg$ continuous in a closed interval $\implies$ $fg$ is uniformly continuous in $A$.
Now Assume $a = \pm \infty$ and/or $b = \pm \infty$. Without loss of generality assume $a = +\infty$.
Now this is the part I'm having a hard time with. If $f,g$ weren't bounded then for example we could have $f(x)=x,\;g(x)=x+1,\;fg(x)=x^2+x$, where we know $f,g$ are uniformly continuous but $fg$ isn't.
So I understand that it $fg$ is uniformly continuous in this situation since $f,g$ are bounded but I'm having a hard time constructing a formal proof.