Stone-Weierstrass theorem for modules of non-self-adjoint subalgebras In "Weierstrass-Stone, the Theorem" by Joao Prolla,  there is a Stone-Weierstrass theorem for modules, stated as the following:

Let $\mathcal{A}$ be a subalegebra of $C(X, \mathbb{R})$ and $(E, \|\cdot\|)$ be a normed space over $\mathbb{R}$. Let $W\subset C(X, E)$ be a vector subspace which is an $\mathcal{A}$-module. For each $f\in C(X, E)$ and $\epsilon>0$, there exists $g\in W$ such that $\|f-g\|<\epsilon$ if and only if for each $x\in X$, there exists $g_x\in W$ such that $\|f(t) - g_x(t)\| < \epsilon$ for all $t\in [x]_{\mathcal{A}}$, where $[x]_\mathcal{A}$ is the equivalent class of $x$ under $\mathcal{A}$.

I know that the above theorem can be extended to $\mathcal{A}\subset C(X, \mathbb{C})$ with $\mathcal{A}$ being a self-adjoint subalgebra. I wonder whether there are some similar results for modules of non-self-adjoint algebras.
I'm interested in generalizing the above theorem into the following case. Let $\mathcal{S}$ be a finite subset of $C([0, 1], E)$, denoted as $S:=\{s_1, \ldots, s_m\}$, and $\mathcal{A}\subset C([0, 1], \mathbb{C})$ be a subalgebra (not necessarily self-adjoint). Then $W := \mathrm{span}\{as : a\in \mathcal{A}, s\in \mathcal{S}\}$ is a vector subspace which is an $\mathcal{A}$-module. Shall we still claim that $f\in \overline{W}$ if and only if $f\big\vert_{[x]_{\mathcal{A}}} \in \overline{W}\big\vert_{[x]_{\mathcal{A}}}$? Is there any counter-example to this statement? Or is it an open problem in general?
Note: For any $x\in X$, the equivalent class $[x]_{\mathcal{A}}$ is a subset of $X$ such that $\forall u, v\in [x]_{\mathcal{A}}$, we have $a(u) = a(v)$ for all $a\in \mathcal{A}$.
 A: If I have understood the definitions correctly, then the answer is still negative, because one can transfer the "disc algebra counterexample" over to $[0,1]$.
In what follows I shall write $C[0,1]$ rather than $C([0,1];{\mathbb C})$, just as a convenient shorthand.
$\newcommand{\cA}{{\mathcal A}}$
$\newcommand{\cB}{{\mathcal B}}$
$\newcommand{\cS}{{\mathcal S}}$
Let $\cB=\{ f\in C[0,1] \colon f(0)=f(1)\}$. For $f\in \cB$ and $n\in \mathbb Z$ let
$$ \widehat{f}(n)= \int_0^1 f(t) e^{-2\pi in t}\,dt $$
(This is the $n$th Fourier coefficient of $f$, if we identify functions in $\cB$ with continuous complex-valued functions on the unit circle in the natural way.) Now let $\cA=\{ f\in \cB \colon \widehat{f}(n)=0\,\forall\,n < 0 \}$. This is a closed subalgebra of $\cB$ and hence a closed subalgebra of $C[0,1]$.
Taking $\cS=\{ {\bf 1} \}$, we have $W=\overline{W}=\cA$.
The equivalence relation on $X=[0,1]$ defined by $\cA$ has the following explicit description: $0\sim_{\cA} 1$; and all other equivalence classes are singletons. This last claim follows by considering the function $t\mapsto e^{2\pi it}$.
In particular, the function $g(t)=e^{-2\pi it}$ belongs to $\cB$ and for every $t\in [0,1]$ we can find $f\in \cA$ such that $f$ agrees with $g$ on $[t]_{\cA}$.
On the other hand, it does not belong to $\cA$, since $\widehat{g}(-1)=1$.
