I am trying to prove the following statement : $(f_n)$ converges uniformly $\iff \sup\{|f_n(x)-f(x)|;\ x\in I\}=\|f_n-f\|_\infty\to 0.$ $\implies$ : We have $\forall \epsilon \leq 0$ $\exists N$ $\forall n \leq N$ $\forall x \in I$ : $|f_n(x)-f(x)| \leq \epsilon$.
We also have that $|f_n(x)-f(x)| \leq \sup\{|f_n(x)-f(x)|;\ x\in I\}$ but then I don't really see how to properly link it to the $\epsilon$.
For $\leftarrow$, I fix an $\epsilon\geq 0$. Then there exists $n_0\in \mathbb{N}$ such that $$\forall n\geq n_0, \sup \{|f(x) - f_n(x)| : x \in I\}\leq \epsilon $$ Which implies, $$\forall n \leq n_0 |f(x) - f_n(x)|\leq \sup \{|f(x) - f_n(x)| : x \in I\}\leq \epsilon , \text{ for any } x\in I.$$ Hence, $f_n\rightarrow f $ as $n\rightarrow\infty$ uniformly. I don't know if I did it correctly though...
$\Longrightarrow:$ Suppose that $(f_n)_{n\in\Bbb N}$ converges uniformly to some function $f$; you want to prove$$\lim_{n\to\infty}\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}=0.\tag1$$Take $\varepsilon>0$. Since $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$, there is some $N\in\Bbb N$ such that$$(\forall x\in I)(\forall n\in\Bbb N):n\geqslant N\implies\bigl|f_n(x)-f(x)\bigr|<\frac\varepsilon2.$$Therefore,$$(\forall n\in\Bbb N).n\geqslant N\implies\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}\leqslant\frac\varepsilon2<\varepsilon,$$and this proves that we indeed have $(1)$.
$\Longleftarrow:$ If we have $(1)$, take $\varepsilon>0$ and take $N\in\Bbb N$ such that$$n\geqslant N\implies\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}<\varepsilon.$$So, if $n\in\Bbb N$ and if $x\in I$, then$$n\geqslant N\implies\bigl|f_n(x)-f(x)\bigr|\leqslant\sup\left\{\bigl|f_n(x)-f(x)\bigr|\,\middle|\,x\in I\right\}<\varepsilon.$$