I do not know much about functional analysis and was hoping to find a reference of the statement that the symmetric power $S^n(C^\infty_c(M))$ of the compactly supported smooth functions on some manifold $M$ is a dense subspace of the symmetric functions $C^\infty_c(M^n)^{S_n}$ on $M^n$. The inclusion $S^n(C^\infty_c(M))\hookrightarrow C^\infty_c(M^n)^{S_n}$ is obtained by
$$(\phi_1\cdots\phi_n)(x_1,\dots,x_n)=\sum_{\sigma\in S_n}\phi_{\sigma(1)}(x_1)\cdots\phi_{\sigma(n)}(x_n).$$
This certainly seems as a version of the Stone-Weierstrass theorem.
Dense subspace of symmetric functions
functional-analysissmooth-functions
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Your idea is good, and you must only take a little more care for it to work. As you said, if $K$ is a compact metric space, then $C(K)$ is separable.
Now, suppose that $X$ is is a locally compact, $\sigma$-compact metric space. You can find a sequence $\left\{K_n\right\}_{n\in\mathbb{N}}$ of compact subsets of $X$ satisfying:
$K_n\subseteq\text{int}K_{n+1}$ for every $n$;
$X=\bigcup_{n\in\mathbb{N}}K_n$.
For every $n$, let $C_n=\left\{f\in C_c(X):\text{supp}f\subseteq K_n\right\}$. Notice that $C_c(X)=\bigcup_{n\in\mathbb{N}}C_n$, so it is sufficient to show that each $C_n$ is separable.
Fixed $n$, consider the function $R_{K_n}:C_n\rightarrow C(K_n)$, $f\mapsto f|_{K_n}$. This is a linear isometry (not necessarily surjective), so $R_{K_n}(C_n)$ is a subspace of the separable space $C(K_n)$, so it is also separable, hence $C_n$ is also separable.
We will need some facts from the convolution theory. Let function $\omega\in C_c^\infty(\mathbb{R}^n)$ such that $\omega\geq 0$ and $\int_{\mathbb{R}^n}\omega(x)\, dx=1$. Denote $\omega_\varepsilon(x):=\varepsilon^{-n}\omega(x/\varepsilon)$.
If $f\in L^1_{\operatorname{loc}}(\mathbb{R}^n)$ then $f*\omega\in C^\infty(\mathbb{R}^n)$ and $D^\alpha (f*\omega)=f*(D^\alpha \omega)$ for all $\alpha\in \mathbb{N}^n_0$.
If $f\in C_c^\infty (\mathbb{R}^n)$ then $\sup\limits_{\mathbb{R}^n}|D^\alpha f-D^\alpha f*\omega_\varepsilon|\to 0$ at $\varepsilon \to 0$ for all $\alpha$ and there are $\varepsilon_0>0$ and compact set $K$ such shat $\operatorname{supp}f*\omega_\varepsilon,\operatorname{supp}f\subset K$, $\varepsilon\leq\varepsilon_0$ (i.e. $f*\omega_\varepsilon \to f$ in $C_c^\infty(\mathbb{R}^n)$).
$\int_{\mathbb{R}^n}|D^\alpha\omega_\varepsilon(x)|\, dx\leq c_\alpha\varepsilon^{-|\alpha|}$.
Fix $f\in C_c^\infty(\mathbb{R}^n)$. We will show that there is $$\{h^\ell\}\subset X:=\operatorname{span}\big(g\, |\, g(x)=g_1(x_1)...g_n(x_n), g_1,...,g_n\in C_c^\infty (\mathbb{R})\big)$$ such that $\sup\limits_{\mathbb{R}^n}|D^\alpha f-D^\alpha h^\ell|\to 0$ for all $\alpha$ and $\operatorname{supp}h^\ell,\operatorname{supp}f\subset K$ for all $\ell$ and for some compact set $K$.
Consider binary cubes $$Q^k_{i_1...i_n}=I^k_{i_1}\times...\times I^k_{i_n},\ k\in\mathbb{N}, i_j\in\mathbb{Z}$$ where $I^k_i=[i/2^k;(i+1)/2^k)$. Let $$f^k(x)=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\chi_{Q^k_{i_1...i_n}}(x)$$ where the number of syllables is finite since $\operatorname{supp}f$ is compact. It is easy to show that $$\delta_k:=\sup\limits_{\mathbb{R}^n}|f^k-f|\to 0.$$
Consider $\omega^1\in C_c^\infty(\mathbb{R})$ such that $\omega^1\geq 0$ and $\int_{\mathbb{R}}\omega^1(x)\, dx=1$. Then nonnegative function $\omega^n\in C_c^\infty(\mathbb{R}^n)$ defined by $$\omega^n(x)=\omega^1(x_1)...\omega^1(x_n)$$ belongs to $C_c^\infty (\mathbb{R}^n)$. Also $\int_{\mathbb{R}^n}\omega^n(x)\, dx=1$ and $\omega^n_\varepsilon(x)=\omega^1_\varepsilon(x_1)...\omega^1_\varepsilon(x_n)$.
I. Using the diagonal Cantor process, we find the sequence $\{\varepsilon(\ell)\}$ such that $\varepsilon(\ell)\to 0$ and $\delta^\alpha_\ell:=\sup\limits_{\mathbb{R}^n}|D^\alpha f-D^\alpha f*\omega^n_{\varepsilon(\ell)}|\to 0$ for all $\alpha$.
II. $f^k*\omega^n_\varepsilon\in X$. Indeed $$f^k*\omega^n_\varepsilon(x)=\int\limits_{\mathbb{R}^n}\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\chi_{Q^k_{i_1...i_n}}(y)\omega^n_\varepsilon(x-y)\, dy$$ $$=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\int\limits_{\mathbb{R}}...\int\limits_{\mathbb{R}}\chi_{I^k_{i_1}}(y_1)...\chi_{I^k_{i_n}}(y_n)\omega^1_\varepsilon(x_1-y_1)...\omega^1_\varepsilon(x_n-y_n)\, dy_1...dy_n$$ $$=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\prod\limits_{j=1}^n\int\limits_{\mathbb{R}}\chi_{I^k_{i_j}}(y)\omega^1_\varepsilon(x_j-y)\, dy=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\prod\limits_{j=1}^n\chi_{I^k_{i_j}}*\omega^1_\varepsilon(x_j)$$ where $\chi_{I^k_{i_j}}*\omega^1_\varepsilon\in C_c^\infty(\mathbb{R})$.
III. We have $$|D^\alpha f^k*\omega^n_\varepsilon(x)-D^\alpha f*\omega^n_\varepsilon(x)|\leq \int\limits_{\mathbb{R}^n}|f^k(y)-f(y)||D^\alpha\omega^n_\varepsilon(x-y)|\, dy\leq \delta_kc_\alpha\varepsilon^{-|\alpha|}.$$
IV. Using the diagonal Cantor process, we find the sequence $\{k(\ell)\}$ such that $\delta_{k(\ell)}\varepsilon(\ell)^{-|\alpha|}\to 0$ for all $\alpha$.
V. Finally let $h^\ell=f^{k(\ell)}*\omega^n_{\varepsilon(\ell)}$. We have $$|D^\alpha f^{k(\ell)}*\omega^n_{\varepsilon(\ell)}(x)-D^\alpha f(x)|\leq |D^\alpha f^{k(\ell)}*\omega^n_{\varepsilon(\ell)}(x)-D^\alpha f*\omega^n_{\varepsilon(\ell)}(x)|+|D^\alpha f*\omega^n_{\varepsilon(\ell)}(x)-D^\alpha f(x)|$$ $$\leq \delta_{k(\ell)}c_\alpha\varepsilon(\ell)^{-|\alpha|}+\delta^\alpha_\ell.$$ I.e. $\sup\limits_{\mathbb{R}^n}|D^\alpha h^\ell-D^\alpha f|\leq \delta_{k(\ell)}c_\alpha\varepsilon(\ell)^{-|\alpha|}+\delta^\alpha_\ell$ where $\delta_{k(\ell)}c_\alpha\varepsilon(\ell)^{-|\alpha|}+\delta^\alpha_\ell\to 0$ for all $\alpha$.
The validity of condition $$\operatorname{supp}h^\ell,\operatorname{supp}f\subset K$$ is easy to check.
Therefore $h^\ell\to f$ in $\mathcal{S}(\mathbb{R}^n)$ (this follows from the properties of the sequence $\{h^\ell\}$, and condition $\operatorname{supp}h^\ell\subset K$ plays an important role).
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Consider the linear map $C_C^{\infty}(M^n)\to C_C^{\infty}(M^n)$ that switches two arguments of a function, eg $$\phi \mapsto [(x_1,..., x_n)\mapsto \phi(x_2,x_1,x_3,...,x_n)].$$ Such a linear map is continuous and as such any permutation of the arguments is a continuous map, further any finite sum of such maps is a continuous linear map. It follows that "symmetrization" $$S:C_C^\infty(M^n)\to C_C^\infty(M^n)^S, \quad \phi\mapsto \frac1{n!}\sum_{\sigma\in S_n}\phi\circ \sigma $$ is continuous (and clearly surjective!), so if $D$ is then a dense subset of $C_C^\infty(M^n)$ then $S(D)$ is dense in $C_C^\infty(M^n)^S$. Now the space $S^n(C_C^{\infty}(M))$ is the space of symmetric tensors, so in order to see that it is dense in the space of symmetric functions it is enough to see that the $n$-fold tensor product of functions is dense in $C_C^{\infty}(M^n)$. This is a standard fact that we will now review.
Let $f\in C_C^{\infty}(M^n)$, by choosing a finite partition of unity of the support of $f$ we may assume that $f$ has its domain entirely in a chart. The question then reduces to whether or not the $n$-fold tensor product of $C_C^\infty(\Bbb R^m)$ is dense in $C_C^\infty((\Bbb R^m)^n)$. For this consider a sequence of polynomials $p_k$ so that $\lim_k \partial_\alpha p_k= \partial_\alpha f$ uniformly in some compact neighbourhood of $\mathrm{supp}(f)$ for all $\alpha$. Let $\psi_1,..,\psi_n$ be smooth bump functions on $\Bbb R^m$ so that
$$\psi_1(x_1)\cdot...\cdot\psi_n(x_n)$$ is equal to $1$ on $\mathrm{supp}(f)$ and $0$ outside of the neighbourhood in which the polynomials converge to $f$. Then $\psi_1...\psi_n p_k$ converges to $f$ in $C^\infty$ sense, further it is clearly a sum of products of functions $\Bbb R^m\to \Bbb C$.