Explanation of an exercise of Hahn-Banach in finite dimensional space The problem is the following:

Let $X$ be a finite-dimensional normed space. Prove that, if $A$ and $B$ are non-empty disjoint convex sets, there exists some hyperplane $H$ that separates $A$ and $B$ (without further assumptions on $A$ and $B$). For proving this, do the following steps:
  
  
*
  
*Consider a non-empty set $C\subset X$ such that $0\not\in C$. Notice that $C$ has a dense and countable subset $\{x_n\}$ [...]
  

The exercise continues but I want to prove the remark I marked in black. My attempt for this is the following.
Let $B=\{b_1,\ldots,b_n\}$, $n\in\mathbb{N}$, a basis for $X$. Then $X=\{x: x=\alpha_i b_i, \alpha_i\in\mathbb{R}\}$. Now, let us consider $S=\{x: x=\beta_i b_i,\beta_i\in\mathbb{Q}\}$ and observe that
$$
\mathbb{Q}\subset\mathbb{R}\Rightarrow S\subset X
$$
and then $S$ can be seen as a countable cartesian product of countables ($S\approx\underbrace{\mathbb{Q}\times\ldots\times\mathbb{Q}}_{n\hspace{0.1cm}\text{times}}$). So $S\subset X$ is countable. Now 
$$
C\subset X\Rightarrow\exists S'\subset S\hspace{0.1cm}\text{such that} \hspace{0.1cm}S'\hspace{0.1cm}\text{is dense in}\hspace{0.1cm} C, S'\subset S\hspace{0.1cm}\text{countable and dense in}\hspace{0.1cm}C.
$$
So $S'=\{x_n\}$, as required. 
My question: is the last statement ($C\subset X\Rightarrow\exists S'\subset S\ldots$) right?
Thanks
 A: As @TrialAndError says, $C$ need not meet $S$ at all. As such your derivation is false. But the statement is true.
First note that in metric spaces the concept of separability and second-countability are equivalent:
If a space is second-countable with a countable open base $\{U_n\}_{n\in \mathbb N}$ then a sequence $\{x_n\}_{n \in \mathbb N}$ where $x_n \in U_n$ is countable. It is also dense since every open set contains at least one of the $U_n$ and thus also one of the $x_n$.
If you have a dense sequence $\{x_n\}_{n \in \mathbb N}$ then the set $\{ B_{1/m}(x_n) \mid n,m \in \mathbb N\}$ is countable. Let $x$ be in $U$ and let $B_\epsilon(x) \subset U$. Since $\{x_n\}$ is dense in every open set contains an element of the set. Let $x_n \in B_{\epsilon/4}(x)$, then from the triangle inequality if $1/m<\epsilon/2$ you have $x\in B_{1/m}(x_n)$ and $B_{1/m}(x_n)\subset B_\epsilon(x) \subset U$. So for any point in $U$ there are sets in the family that contain the point and are contained in $U$. This makes it a base.
In a topological space it follows directly from the definition of the subset topology that second-countability is conserved in restricting to a subspace.
This means that every subset of $X$ is second-countable and thus separable if $X$ is separable and metric.
