uncountable subset of $C[0,1]$ has uniformly convergent subsequence If $S$ is an uncountable subset of $C[0,1]$, then there is a uniformly convergent sequence $\{f_n\}$ of distinct functions of $S$. 
I know how to do this for $C^1[0,1]$ since $S \subset 
\cup_{m,n \ge 0} \{f: \sup_{[0,1]} |f(x)| \le m,  \sup_{[0,1]} |f'(x)| \le n \}$ 
and then $\{f: \sup_{[0,1]} |f(x)| \le m,  \sup_{[0,1]} |f'(x)| \le n \}$ must be uncountable for some $m,n$ (since otherwise $S$ would be countable) and so contains a uniformly convergent sequence by Arzelà–Ascoli.

Suppose $S$ is such that no $x \in C[0,1]$ is the uniform - that is normwise - limit of a sequence $(x_n)$ of distinct elements of $S$.
Let $x \in C[0,1]$, then there is an $\epsilon_x > 0$ such that $U_{\epsilon_x}(x)= \{y \in C\in C[0,1] \mid \|x-y\| < \epsilon\}$ has finite intersection with $S$ (otherwise, for $n \in\mathbb N$ let $x_n \in U_{1/n}(x) \cap S$ pairwise distinct, giving $x_n \to x$). We have $C[0,1] = \bigcup_x U_{\epsilon_x}(x)$. As $C[0,1]$ is seperable metric, it is Lindelöf, therefore there is a countable $A \subseteq C[0,1]$ with $C[0,1] = \bigcup_{x\in A} U_{\epsilon_x}(x)$. Now $S = \bigcup_{x \in A} (U_{\epsilon_x}(x) \cap S)$ is a countable union of finite sets, hence countable. 
As your $S$ is uncountable, there is a normwise convergent sequence $(x_n)$ of distinct elements of $S$.
