EDIT: answer provided. Just a slip up confusing the order of taking $sup$.
EDIT to EDIT: While I did make the conceptual mistake above, turns out my conclusion was not incorrect.
I have a function $f(x,y)$ and want to show that for some $C>1$, $C = \underset{x,y}{sup}f(x,y)$. I know also that for any $x,y$, I have $f(x,y) \leq C$.
I could go about this by saying "take an arbitrary $A$ such that $1 < A < C $. If I can show that $ \underset{x,y}{sup}f(x,y) > A$, then I am done".
Now apparently the strict inequality is necessary in the above. But I am struggling to understand why. Why is it not enough to show that $f(x,y) \geq A$ for arbitrary such $A$? Since could I not then bring $A$ arbitrarily close to $C$ anyway, so the supremum of $f(x,y)$ would have to be $C$?
I think You are right : suppose You only showed that $sup_{x,y}f(x,y)\geq A$ for arbirtary $A<C$. You know $sup_{x,y}f(x,y)\leq C$ and now You suppose $sup_{x,y}f(x,y)<C$, then there exists a number $A$ such that $$sup_{x,y}f(x,y)<A<C$$ but this contradicts $sup_{x,y}f(x,y)\geq A$ for every $A<C$ and thus $sup_{x,y}f(x,y)=C$.