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A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzela-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

Fix $x$ and $\epsilon >0$. Since $\mathcal{K}$ is compact in the uniform norm, choose a finite $\epsilon /3$-net $f_1,\dots ,f_r$. By continuity, there is a neighbourhood $U$ of $x$ on which $|f_i(y)-f_i(x)|<\epsilon /3$ for every $i$. For $f$ in $\mathcal{K}$, choose $f_i$ with $\Vert f-f_i{\Vert }_{\infty }<\epsilon /3$. The triangle inequality gives $|f(y)-f(x)|<\epsilon$ on $U$, uniformly in $f$. Thus $\mathcal{K}$ is equicontinuous.

The sequence $(f_n)$ is equicontinuous on $[0,1]$ when, for every $\epsilon >0$, there is a $\delta >0$ such thatfor every $n$. The Arzela-Ascoli theorem says that a uniformly bounded equicontinuous sequence of continuous real-valued functions on a compact interval has a uniformly convergent subsequence.