Show that any compact subset of $C([0,1])$ with sup norm is equicontinuous

Question

Question is stated as follows: Show that any compact subset of $C([0,1])$ with the sup norm $\lVert \cdot\rVert_{\infty}$ is equicontinuous.

Answer 1

Let $F$ be a compact subset of $C[0,1]$. Let $x \in [0,1]$ and fix $\epsilon > 0$. For $n \in \mathbb{N}$, define $$ A_n := \left\{f \in C[0,1]: \sup_{|x-y| \leq 1/n } |f(x)-f(y)| < \epsilon \right\}$$ Then, for every $n \in \mathbb{N}$, the set $A_n$ is open, and furthermore $$F \subseteq C[0,1] \subseteq \bigcup_{n\in\mathbb{N}} A_n.$$ By compactness, there is a finite number of $n_1,\dots,n_k \in \mathbb{N}$ such that $$F \subseteq A_{n_1} \cup \dots \cup A_{n_k}.$$ Let $l \geq \text{max}(n_1,\dots,n_k)$, then $F \subseteq A_{l}$, which proves that $F$ is equicontinuous.

To prove that the $A_n$ are open, let $f \in A_n$ and take $g$ such that $$2\|f-g\| < \epsilon - \sup_{|x-y| \leq 1/n } |f(x)-f(y)|.$$ Then, \begin{align} \sup_{|x-y| \leq 1/n } |g(x)-g(y)| &\leq \sup_{|x-y| \leq 1/n } |g(x) - g(y) - f(x)+ f(y)| + \sup_{|x-y| \leq 1/n } |f(x)-f(y)| \\ & \leq 2 \|f-g\| + \sup_{|x-y| \leq 1/n } |f(x)-f(y)| \\ &< \epsilon, \end{align} so $g \in A_n$ and hence $A_n$ is open.