In an attempt to better understand the definition of an equicontinuous family of continuous functions, I want to find a simple non-example.
My intuition says that the family $\{f_n\colon[0,1]\to\mathbb R\}_{n\in\mathbb N}$ given by $f_n(x)=x^n$ is not equicontinuous, but I do not know how to show this.
Any help is appreciated.
We claim that the point $x=1$ ruins the family's chance at equicontinuity. We choose $\epsilon = \frac{1}{2}$ and let $\delta>0$ be given. Pick any $y\in(1-\delta,1)$ you want. Letting $n\to \infty$ we have $y^n \to 0,$ so we may choose an $n$ large enough that $y^n < \frac{1}{2}$, then we would have $$ |f_n(1)-f_n(y)| = |1-y^n| \geq \frac{1}{2}. $$
Hence the family is not equicontinuous at $x=1$.
According to Wikipedia (link given in OP): "...the limit of an equicontinuous pointwise convergent sequence of continuous functions $f_n$ on either metric space or locally compact space is continuous. ..."
In your example the pointwise limit of $f_n$ is not continuous. It is $0$ everywhere except at $1$ (where it's $1$). So we may conclude that $f_n$ are not equicontinuous.
The problem is at $1$. Indeed, assume that $\{f_n\}$ is equi-continuous at $1$. Then we can find a $\delta>0$ such that if $|1-x|\leq \delta$ and $0\leq x\leq 1$ then for each integer $n$, $|f_n(x)-f_n(1))|\leq 1/3$, hence $|x^n-1|\leq 1/3$ and $x^n\geq 2/3$. If $1/{n^2}\leq \delta$ then we should have $\left(1-\frac 1{n^2}\right)^n\geq 2/3$. It's impossible, since the RHS converges to $0$.
In fact, we know by Arzela-Ascoli-'s theorem that a uniformly bounded sequence of equi-continuous functions on a compact interval admit a converging subsequence for the uniform norm (and this limit is continuous). We can check that the pointwise limit of $\{f_n\}$ is not continuous.
