Let $E$ a topological vector space over $\mathbb{R}$ or $\mathbb{C}$, and $S \subset E$. Then $\operatorname{span}_{\mathbb{Q}} S$ is dense in $\operatorname{span}_{\mathbb{R}} S$, and if we have a complex vector space, $\operatorname{span}_{\mathbb{Q}(i)} S$ is dense in $\operatorname{span}_{\mathbb{C}} S$. These follow immediately from the denseness of $\mathbb{Q}$ in $\mathbb{R}$ and of $\mathbb{Q}(i)$ in $\mathbb{C}$ together with the continuity of the vector space operations.
Let's assume the complex case to save typing alternatives.
Since $\mathbb{Q}(i)$ is countable, $\operatorname{span}_{\mathbb{Q}(i)} S$ is countable if $S$ is countable.
Now let $M\subset E$ be a separable subset, and $D\subset M$ a countable dense subset. Then $F := \overline{\operatorname{span}_{\mathbb{Q}(i)} D}$ is the closure of a countable set, hence separable. Further,
$$\operatorname{span}_{\mathbb{C}} D \subset F$$
by the above, and since $\operatorname{span}_{\mathbb{Q}(i)} D \subset \operatorname{span}_{\mathbb{C}} D$, we have
$$F = \overline{\operatorname{span}_{\mathbb{C}} D},$$
so $F$ is in fact a closed linear subspace of $E$. Since trivially $M \subset \overline{D} \subset \overline{\operatorname{span}_{\mathbb{Q}(i)} D}$, $F$ is a subspace containing $M$, and hence $\operatorname{span}_{\mathbb{C}} M \subset F$. Thus $\operatorname{span}_{\mathbb{Q}(i)} D$ is a countable dense subset of $\operatorname{span}_{\mathbb{C}} M$.
