# 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.

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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 $S$.
@joriki And (norm-)convergence in $C[0,1]$ is uniform convergence, right? – martini Sep 20 '12 at 8:02