I want to prove that for a bounded function $f$, we have:
$$ ||f||_\infty =0 \implies f=0. $$
Simply observe that if we denote the domain of definition by $D$: $$ ||f||_\infty = \sup_{x \in D}|f(x)|$$ We know that the supremum is the least upper bound, so for any $y \in D $: $$ |f(y)| \leq \sup_{x \in D} |f(x)|=0$$ But we also know that the Euclidean metric is positive definite, so $|f(y) |\geq 0$, and we get that for all $y \in D$ we have $|f(y)|=0$, but now again by positive definiteness:
$$|f(y)|=0 \iff f(y)=0.$$
Is my reasoning correct?
Yes it is correct, alternatively, as suggested by Quiliup, We can immediately conclude that since for all $y \in D$ $$|f(y)|\leq 0$$ By positive definiteness of the Euclidean norm it must hold that for all $y \in D$ we have $|f(y)|=0$ and hence $f(y)=0$.