Let $\{f_n:E\rightarrow \mathbb{R}\}$ be a sequence of functions which uniformly converges to $f$ in $E$.
Prove: $$\sup_Ef_n \rightarrow \sup_Ef$$
Proof:
By definition of supremum: $$f_n(x)-f(x)\leq \sup_E|f_n-f|, \ \forall x\in E$$
Rearranging:
$$f_n(x)\leq \sup_E|f_n-f| +f(x)$$
Taking supremum on both sides:
$$\sup_Ef_n(x)\leq \sup_E|f_n-f| +\sup_Ef(x)$$
In symmetric way, we get:
$$\sup_Ef(x)\leq \sup_E|f_n-f| +\sup_Ef_n(x)$$
All together:
$$|\sup_Ef_n(x)- \sup_Ef(x)| \leq \sup_E|f_n-f| \rightarrow 0 $$
I'm not sure about the step I took supremum on both sides, so if someone could please verify / correct my proof, I would appreciate it.