Is there an alternative way to show that a finite dimensional vector space is separable?

Question

I am aware of the proofs of properties of finite-dimensional vector spaces in functional analysis. These use the properties of normed spaces $(\mathbb R^{d},\lvert\lvert\cdot\rvert\rvert)$ and establish an isometric isomorphism to more general $d-$dimensional vector spaces. Properties such as completeness, closedness and separability immediately follow.

Particularly on the issue of separability, is there any other way to show that a finite-dimensional normed space is indeed separable without using this isometry?

Answer 1

Let $\{x_1,x_2,..,x_n\}$ be a basis. Then $\|\sum a_i x_i-\sum b_ix_i\| \leq \max_i |a_i-b_i| \sum \|x_i\|$. From this it is clear that rational linear combinations of $x_i$'s form a countable dense subset.

Answer 2

If $X$ is finite dimensional then it has a finite basis $\{x_1,...,x_n\}$. Then you can check that the linear combinations of $x_1,...,x_n$ with coefficients from $\mathbb{Q}+i\mathbb{Q}$ is a countable dense subset.