Let $K$ a separable, compact metric space. We can say that $C(K,\Bbb R)$ is also separable without more assumptions over $K$?
This question is mine, I mean, it doesn't come from some exercise or so and Im not sure if it have a positive answer.
If $K$ is a finite-dimensional subset of a vector space it would be easy to show, using the Stone-Weierstrass theorem and polynomials of rational coefficients, that $C(K,\Bbb R)$ is separable.
However, we can say the same without more assumptions over $K$ that it is compact, separable and metric?
This essentially follows from a form of Stone-Weierstraß (if $A$ is a unital subalgebra of $C(K)$, with $K$ compact Hausdorff, that separates points, then $\overline{A} = C(K)$, which is a well-known standard form):
If $(K,d)$ is compact metric, it's separable so pick a countable dense $D = \{d_n: n \in \mathbb{N}\}$ and let $A$ be the unital subalgebra generated by the countably many functions $f_{n}(x) = d(x,d_n)$. These separate the points of $X$. Hence $A$ is dense (and has a countable dense subset itself in the rational-linear combinations of the generating functions etc.)