Does uniform continuity on a compact subset imply equicontinuity?

Question

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?

Answer 1

To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.

An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $\delta > 0$ such that if $|x - 1| < \delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n \to 0$ when $1- \delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.