I’m trying to prove that if $f_n \to f$ uniformly, where each $f_n$ is bounded and defined on $E\subset \mathbb{R}$, then there exists $A$ such that $f_n(x)\leq A$ for all $n$. I came up with a proof that involves the supremum norm, and I wanted to make sure it’s correct since most solutions take a different approach.
Let $A_n= \sup_{x\in E} |f_n(x)|=\|f_n\|$. Then, by uniform convergence, $\|f_n-f\| \to 0$ (in baby Rudin, this is Theorem 7.9). Since the supremum norm is a norm, the reverse triangle inequality holds:
$$ \|f_n-f\| \geq \big|\|f\| - \|f_n\|\big| = \big|\|f\| - A_n \big|\geq 0$$
Hence, $\big|\|f\| - A_n \big|\to 0$ by the squeeze theorem, so in particular $A_n$ is a bounded sequence, so we may take $A=\sup_n A_n$ to ensure $|f_n(x)|\leq A$ for all $n$.