I gave a proof here some time ago but was wondering about the following:
Suppose that $f_n:K\subseteq \Bbb{R}\to\Bbb{R}$ is continuous for each $n\in \Bbb{N}$, then $\{f_n\}$ is uniformly continuous. So, for every $\epsilon > 0$ and $n\in \Bbb{N}$, there exists $\delta_n>0$ such that $\forall\,x,y\in K, \;|x − y| < \delta_n,$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}.$$ Taking $\delta=\min\{\delta_n:\;n\in \Bbb{N}\}$ (which might not be true), then we have the definition given here, that "for every $\epsilon > 0$, there exists $\delta>0$ such that $\forall\,x,y\in K, \;|x − y| < \delta,$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}.$$ which implies that $\{f_n\}$ is equicontinuous.
QUESTION: I'm I right? If not, can you please provide a counter-example?