Let's say $A$ is compact and $f,g$ are continuous. Is it true that $$[\sup\limits_{x\in A}(f(x)-g(x))][\sup\limits_{x\in A}(f(x)+g(x))]=\sup\limits_{x\in A}(f(x)^2-g(x)^2)?$$
All my intuition say yes, but I am concerned that there is something more complicated going on here given the supremum.
It is not true in general that $\sup_xa(x)\sup_xb(x)=\sup_xa(x)b(x)$, which is a generalization of what you're saying here. For example, if $a$ and $b$ are indicator functions of disjoint sets, then $ab$ is identically zero but $\sup a$ and $\sup b$ are both $1$.
Any such example can be adapted as a counterexample to your claim simply by finding $f$ and $g$ such that $f-g=a$ and $f+g=b$.
Another simple counterexample would be to set $g=0$, at which point you're claiming $(\sup f)^2=\sup(f^2)$, which is false for example for a function $f$ which takes on values of $-1$ and $0$.
All of these examples can be made continuous by interpolation without changing the conclusions.
As mentioned in the comments, the basic problem is that in $\sup a\sup b$, we're free to take the largest value of $a$ and the largest value of $b$ separately. In order for that to equal $\sup ab$, those values would have to occur at the same value of $x$.