Continuity of a function $g(x)$ at $x=1$ where $(f(x) cdot g(x))'$ exist at $x=1$

Let $;g;$ be the most non-continuous function at $;x=1;$ but bounded in some neighborhood of that point...For example, take

$$g(x)=begin{cases}1,&x=1\{}\0,&text{otherwise}end{cases}implies h(x)=f(x)g(x)=0$$

and thus $;h(x);$ is infinitely differentiable everywhere, yet $;g;$ isn't continuous at $;x=1;$

What if $g(x)$ is a function returning zero for every input except $x = 1$, where it returns $1$? Then $h(x) = 0$ for all $x$ and is clearly differentiable, but $g(x)$ is clearly not continuous at $x = 1$.

I don't quite understand your attempted proof unfortunately. First, the product rule is used to calculate the derivative of a product of two differentiable functions. If $f$ and $g$ aren't necessarily differentiable, then you can't apply it here. Second, even the conclusion of the proof does not necessarily show it's continuous. Instead, you want to show that as $x to 1$, $g(x) to g(1)$.