This is just a summary.
Theorem
Given a compact domain $\Omega$.
Regard the function space: $$\mathcal{C}(\Omega,\mathbb{R}):=\{f:\Omega\to\mathbb{R}:f^{-1}\mathcal{T}(\mathbb{R})\subseteq\mathcal{T}(\Omega)\}$$
Denote for shorthand: $$\mathcal{F}\subseteq\mathcal{C}(\Omega,\mathbb{R}):\quad\mathcal{F}[x]:=\{f(x):f\in\mathcal{F}\}$$
Then separating algebras with identity are dense: $$1\in\mathcal{A}\leq\mathcal{C}(\Omega,\mathbb{R}):\quad\mathcal{A}[x]\neq\mathcal{A}[y]\quad(x\neq y)\implies\overline{\mathcal{A}}=\mathcal{C}(\Omega,\mathbb{R})$$ This follows from the lemma and the proposition below.
Remark
How does this extend to the complex version?
Corollary
Polynomials on a compact euclidean domain are dense!
Proposition
Consider a lattice that satisfies: $$\mathcal{L}\preccurlyeq\mathcal{L}(\Omega,\mathbb{R}):\quad l\in\mathcal{L}\implies \lambda l,\lambda+l\in\mathcal{L}$$
Then a separating one was already dense: $$\mathcal{L}[x]\neq\mathcal{L}[y]\quad(x\neq y)\implies\overline{\mathcal{L}}=\mathcal{C}(\Omega,\mathbb{R})$$ That is a check step by step.
Lemma
The modulus can be approximated by polynomials: $$M(t):=|t|:\quad\|M-P_n\|_{[-1,1]}\to0\quad(P_n\in\mathcal{P}[-1,1])$$ Alternatively one can approximate the square root: $$R(s):=\sqrt{1-s}:\quad\|R-Q_n\|_{[0,1]}\to0\quad(Q_n\in\mathcal{P}[0,1])$$ (For a detailed check see: Modulus)
Proof
Clearly, all algebraic properties remain: $$1\in\overline{\mathcal{A}}\leq\mathcal{C}(\Omega,\mathbb{R}):\quad \overline{\mathcal{A}}[x]\neq\overline{\mathcal{A}}[y]\quad(x\neq y)$$
In particular, it is a lattice then since: $$f\neq0:\quad f\in\overline{\mathcal{A}}\implies P_n(f)\in\overline{\mathcal{A}}\implies|f|\in\overline{\mathcal{A}}\quad(\|f\|\leq1)$$ That proves the assertion.
Credits
Thanks alot for suggesting me Royden, T.A.E.!!
Disclaimer
This thread is meant to record. See: Answer own Question
This answer is community wiki.
Suppose it is additional selfadjoint: $$\mathcal{A}\leq\mathcal{C}(\Omega,\mathbb{C}):\quad\mathcal{A}^*=\mathcal{A}$$
Reduce to the real case by: $$1\in\mathcal{A}_\mathbb{R}:=\{g\in\mathcal{A}:g=g^*\}\leq\mathcal{C}(\Omega,\mathbb{R})$$
This remains separating as: $$g_x(x)\neq g_y(y)\quad(x\neq y)\implies\Re{g_x}(x)\neq\Re{g_y}(y)\lor\Im{g_x}(x)\neq\Im{g_y}(y)$$
So applying the real version gives: $$g\in\mathcal{C}(\Omega,\mathbb{C}):\quad f_n+if'_n\to\Re{g}+i\Im{g}=g\quad(f_n,f'_n\in\mathcal{A}_\mathbb{R})$$
That extends the above result!