Let $X$ be a compact metric space. Let $F\subseteq C(X;\mathbb{R})$ be a family of functions. Then according to Arzela Ascoli $F$ is compact if and only if $F$ is equicontinuous at any point $t$ and $F$ is uniformly bounded.
But I think uniform boundedness is implied already by equicontinuity.
So equicontinuity should be equivalent to compactness here, no?
Because of $\sup\limits_{f\in F}|f(x)-f(y)|\rightarrow 0$ $(x\rightarrow y)$
one has
$\left|\sup\limits_{f\in F}|f(x)|-\sup\limits_{f\in F}|f(y)|\right|\rightarrow 0$ $(x\rightarrow y)$
So $\sup\limits_{f\in F}|f(x)|$ is continuous and since $X$ is compact one has
$\sup\limits_{x\in X}\sup\limits_{f\in F}|f(x)|<+\infty$
but $\sup\limits_{x\in X}\sup\limits_{f\in F}|f(x)|=\sup\limits_{f\in F}\sup\limits_{x\in X}|f(x)|$
so $f$ is uniformly bounded.
A collection of constant functions $f_n(x) = n$ for all $x$ is equicontinuous but not uniformly bounded.