 Solution

A normed vector space is a real or complex vector space $X$ with a norm satisfying positivity, definiteness, absolute homogeneity and the triangle inequality. It is separable if it has a countable dense subset for the norm distance.

First prove the unlabelled general assertions. If $(x_n)$ is dense in $X$, choose $y_{n,m}\in Y\cap B(x_n,1/m)$ whenever that intersection is nonempty. These choices form a countable subset of $Y$. For any $y\in Y$ and $\epsilon >0$, choose $m$ with $2/m<\epsilon$ and $n$ with $\Vert x_n-y\Vert <1/m$. The corresponding intersection is nonempty and $\Vert y-y_{n,m}\Vert <2/m<\epsilon$. Thus every subspace of a separable normed vector space is separable.

If a countable $S$ has dense linear span, take all finite linear combinations of its elements with rational coefficients in the real case, or coefficients in $\mathbb{Q}+i\mathbb{Q}$ in the complex case. This set is countable and dense in the linear span, by approximating the finitely many coefficients of each combination. Its closure is therefore $X$, proving separability. Conversely, a countable dense set already has dense linear span.

For $c_0$, use all finitely supported sequences with coordinates in $\mathbb{Q}+i\mathbb{Q}$. Truncate a sequence tending to zero to control its supremum tail, then approximate the finite initial coordinates. This produces a countable dense subset, so