$(X,d)$ separable implies uniformly equicontinuous subset of $C(X,\mathbb{R})$ is separable?

Question

I'm trying to prove a certain class of subsets of $C(X)$ (made into a metric space, equipped with the sup-norm) are separable iff $X$ is (where $X$ is a metric space) and while one direction is easy, it's much tougher to prove that the subset of $C(X)$ is separable if $X$ is.

I have taken a basic course in topology and read a fair amount, so I've been trying to use/generalise a version of Arzela-Ascoli (which assumes $X$ is compact and various other stronger conditions) that we proved in the course. In particular, I know that the subspaces of $C(X)$ I'm considering are uniformly equicontinuous (indeed they are 1-Lipschitz) and I was hoping that this would be enough to deduce separability of the subset.

I know there are a few steps in typical Arzela-Ascoli-type proofs:

1. You require the input space to be separable to get a countable dense subset (and this is what we are assuming!).
2. You show that given some (uniformly) bounded sequence of functions has a "diagonal" subsequence such that the members' outputs converge on the dense subset (this makes no assumptions on $X$).
3. You show that if $(f_n)$ is uniformly equicontinuous and converges on the dense set, then it converges in the maximum norm to some $f$. (The proof I know only uses $X$ being totally bounded for this step.)
4. You put this altogether, along with maybe some fact like compact iff closed and totally bounded, to deduce your Arzela-Ascoli.

I'm not really sure how to adjust steps 3 and 4 to deduce separability...

EDIT:

I've found an essentially unrelated argument which directly shows that if $S\subseteq C(X)$ is bounded and uniformly equicontinuous and $X$ is totally bounded, then $S$ is totally bounded. This is almost but not quite what I want!

Answer 1

Your question is a bit mysterious so I don't precisely know what are your assumptions on the subspace. Nevertheless I think it isn't true.

Counterexample Let $X=[0,\infty)$. It's separable. For any $c\in C:=\{0,1\}^{\Bbb N}$ there is a $1$-Lipschitz function $f_c\in C(X)$ such that $f_c(n)=c(n)$. Observe that if $c_1\neq c_2$ then $\|f_{c_1}-f_{c_2}\|_{\sup}\geq 1$. Then $F=\{f_c:c\in C\}$ is an uncountable discrete subset of $C(X)$. Therefore it isn't separable.