Cardinality of a special set of continuous functions Let $C$ be a set of continuous functions with a domain $[0,1]$ and for every input $x$ in a domain there is a set $S(x)$ that contains all values that functions in $C$ will output given that input.
For example, if $f$, $g$ and $h$ are all functions in $C$, than, for example $f(0.5)$, $g(0.5)$ and $h(0.5)$ are all in $S(0.5)$. Of course, if these are the only functions in $C$ and all three outputs $f(0.5)$, $g(0.5)$ and $h(0.5)$ are different, than cardinality of $S(0.5)$ is 3. If they all are the same, than cardinality of $S(0.5)$ is 1.
The question is, given the fact that $card(S(x))= \aleph_0$ for all $x$ in a domain, can there be such $C$ that $card(C)> \aleph_0$ ?
What about the case when $C$ is a set of smooth functions? Analytic?
 A: For your initial question $-$ yes, we can find such $C$ as Saúl Rodríguez Martin showed in their comment.
For the analytic functions, I believe it is exactly Wetzel's problem: the solution is independent from ZFC. But with continuum hypothesis being true, we can find such $C$.
For smooth functions I didn't find solution, but here is an observation. If there is $C$ satisfying your conditions and contains only functions that can be differentiated at most $N$ times, than there surely is satisfying $C$ that with functions differentiable $N-1$ times as it is less restrictive. So if there is some $N$ for which there is no $C$, than there are no solutions above $N$ (because if there is solution for $N+1$ than there is for $N$).
A: Original problem with continuous functions: A counterexample is given by the set $C$ of continuous functions $f$ obtained by setting $f(1)=0$, $f(1-2^{-n})\in\{0,2^{-n}\}$ and interpolating linearly.
$C^\infty$ functions:
Let $\phi:\mathbb{R}\to\mathbb{R}$ be a $C^\infty$ bump function with support $[-1,1]$, and $\phi(p,r)(x)=\phi\left(\frac{x-p}{r}\right)$ be the same bump function but centered in $p$ and with support of radius $r$.
Now for a given positive sequence $\mathbf{x}=(x_n)_n$ let $F_\mathbf{x}=\sum_{n\in\mathbb{N}}x_n\phi(1-3\cdot2^{-n},2^{-n-2})$. Then for any $f\in C$, $f\cdot F_\mathbf{x}$ is $C^\infty$ in $[0,1)$. Moreover, if we make the $x_n$ decrease fast enough, we can ensure that $\forall f\in C$, $\lim_{x\to1}(f\cdot F_\mathbf{x})^{n)}(x)=0$ $\forall n$, so that $f\cdot F_\mathbf{x}$ is $C^\infty$ in $[0,1]$. So $\{f\cdot F_\mathbf{x};f\in C\}$ does the job.
Analytic functions: In this case the existence of $\mathcal{C}$ is equivalent to the continuum hypothesis.
If the continuous hypothesis is true, let $\omega_1$ be the set of countable ordinals and consider a numbering $(x_\alpha)_{\alpha\in\omega_1}$ of $[0,1]$. We can create recursively functions $\{f_\alpha\}_{\alpha\in\omega_1}$ which only take countably many values at any point: it is enough to, for each ordinal $\alpha$, let $f_\alpha$ be a function which:

*

*For every $\beta<\alpha$, $f_\alpha(x_\beta)$ is rational.

*$f_\alpha$ is distinct from $f_\beta$ $\forall\beta<\alpha$.

It is possible to create such $f_\alpha$ due to the following lemma:
Lemma: Given a countable set $(x_n)_{n\in\mathbb{N}}$ in $[0,1]$ and some $y_0\in\mathbb{R}$ we can create an analytic function $f:[0,1]\to\mathbb{R}$ with $f(x_0)=y_0$ and $f(x_n)\in\mathbb{Q}$ $\forall n$
Proof of the lemma: We can define $f=y_0+\sum_{n\geq1}P_n$, where $P_n$ are polynomials such that:

*

*$P_n(x_m)=0\;\forall m<n$.

*$P_n(x_n)$ is adjusted so that $f(x_n)\in\mathbb{Q}.$

*The $P_n$ decrease very fast with $n$ so that $f$ is analytic in $[0,1].\square$
Then the $f_\alpha$ are an uncountable collection of analytic functions, but $S(x)$ is countable $\forall x\in[0,1]$.
If the continuum hypothesis is false, suppose you have an uncountable family of analytic functions $(f_\alpha)_{\alpha\in\omega_1}$. Notice that any two distinct analytic functions can only coincide in countably many points. So the set $\{x\in[0,1];f_\alpha(x)=f_\beta(x)\text{ for some }\alpha\neq\beta\}$ has cardinal at most $\aleph_1<2^{\aleph_0}$, and for any $x$ outside of that set, $S(x)$ is uncountable.
Edit: As user479568 mentions in his answer, the analytic case is Wetzel's problem, and Erdös gave a solution in this paper (although it is almost the same as the solution I give here).
