= 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 $\varepsilon>0$, choose $m$ with $2/m<\varepsilon$ and $n$ with $\|x_n-y\|<1/m$. The corresponding intersection is nonempty and $\|y-y_{n,m}\|<2/m<\varepsilon$. Thus \b[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
$$
\boxed{c_0\text{ is separable}.}
$$
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