prove that $D=\overline{\text{span}\{x_n:n \in \mathbb{N}\}}$ is separable Let $E$ be a normed space. Let $\{x_n: n \in \mathbb{N}\} \subset E$. 
I need to prove that $D=\overline{\text{span}\{x_n:n \in \mathbb{N}\}}$ is separable.
I've defined $A_n= \left\{ \displaystyle\sum_{j=1}^n a_j x_j : a_j \in \mathbb{Q} \right\}$ and $A= \displaystyle \bigcup_{n=1}^{\infty} A_n$. 
I don't know how to prove that $A$ is dense, and I don't know exactly why $A_k$ is countable.
I need help. Thanks.
 A: A solution follows, but I encourage you to think about these hints first:

*

*approximate $y \in D$ by $x$ in the span of the sequence. Now approximate $x$ by something in $A$. By taking both of these near enough, one should obtain an element of $A$ as near to $y$ as desired.

*union of countable sets are countable, so it suffices to show that each $A_n$ is countable. Think about why finite rational sequences are countable.

Let's see that $A$ is dense in $D$. Pick $y \in D$ and $\varepsilon > 0$. By definition of closure, you can take $x \in \langle x_n \rangle_{n \geq 1}$ such that $\|y-x\| < \varepsilon/2$. Now, the element $x$ must be a finite linear combination of terms of this sequence. There must exist then $a_1, \dots, a_n \in \mathbb{R}$ such that
$$
x = a_1x_1 + \cdots a_nx_n. \tag{1}
$$
Think about why this is true: surely $x$ is a linear combination of some terms, so you can "fill the gaps" between these by choosing $a_i = 0$ obtaining $(1)$.
Finally, by density of the rationals, pick rationals $q_1,\dots,q_n$ such that $|q_i-a_i|\|x_i\| < \varepsilon/2n$ so that noting $z = \sum_{i=1}^nq_ix_i$ we have
$$
\|x-z\| = \left\|\sum_{i=1}^n(a_i-q_i)x_i\right\| \leq \sum_{i=1}^n|a_i-q_i|\|x_i\| < \varepsilon/2.
$$
Therefore $z \in A$ and $\|y-z\| \leq \|y-x\| + \|x-z\| < \varepsilon$.
Now let's show that $\#A_n$ is countable. Recall that if $X$ is countable, so is $X^n$. Finally, note that the assignment
$$
(q_1,\dots,q_n) \in \mathbb{Q}^n \mapsto \sum_{i=1}^nq_ix_i \in A_n
$$
is surjective, and so $\aleph_0 = \#\mathbb{Q}^n \geq \#A_n$.
