Normed vector spaces and separability This question was asked in my assignment of functional analysis and I am not able to make much progress on this issue.

Question:  Let X be a normed vector space. Show that (a) is equivalent to (b):

(a) X is separable.
(b)  there exists an increasing sequence $\cup_{n\in \mathbb{N}}E_n$ of finite dimensional subspaces of X such that $\cup_{n\in \mathbb{N}}E_n $ is dense in X.
Attempt: (a) to (b)  I am not able to make any reasonable progress.
(b)=>(a) consider the set $\cup_{n\in \mathbb{N}} E_n$ . This set is dense in X. I have to show that this set is also countable. $E_n$ is a sequence of finite dimensional subspaces which are countable. So, the sequence is countable but I have to show that all the elements in the set $\cup_{n\in \mathbb{N}} E_n$. are countable. But I am not able to show this using the fact that sequence of sets $E_n$ is countable. I have to use some property of counting from elementary set
theory but I am unable to. (I think the cardinality of the set $\cup_{n\in \mathbb{N}} E_n$  is finite as it will always be less than or equal to the cardinality of $\mathbb{N} \times \mathbb{N}$ and the latter is countable. Is this argument fine?)
Kindly help me with this!
 A: If $E$ is a normed vector space, then there is one, and only one, linear subspace of $E$ that has the property of being countable: $\{0\}$.
Any other linear subspace contains a line $\Bbb R v$ with $v\neq 0$, which has the cardinality of $\Bbb R$: it cannot be countable.
However, a finite dimensional normed vector space is always separable.
Indeed, let $E$ be such a finite dimensional vector space, and $\{v_1,\cdots,v_N\}$ be a finite generating family.
Then $F=\Bbb Q v_1 + \cdots + \Bbb Q v_N$ is a dense countable subset.

Getting back to your question:

*

*$(a) \implies (b)$.
Assume that $X$ is countable, and let $\{x_1,\ldots,x_n,\ldots\}$ be a countable dense sequence in $X$, and for $n\geqslant 1$, define $E_n= \mathrm{span}\{x_1,\ldots,x_n\}$.
By construction, $(E_n)$ is an increasing sequence of finite dimensional subspaces.
In addition, $\{x_0,\ldots,x_n,\ldots\} \subset \cup_{n} E_n \subset X$, and it follows that $\cup_nE_n$ is dense in $X$.


*$(b)\implies (a)$. From what I said in the introduction of my answer, there exists $F_n\subset E_n$ a dense countable subset in $E_n$.
Define $F=\cup_n F_n\subset X$.
Then $F$ is countable as a countable union of countable sets.
Let us show that $F$ is dense in $X$.
This is equivalent to showing that for any non-empy open set $U$, $U\cap F\neq \varnothing$.
Let $U$ be such a non-empy set.
Since $\cup_nE_n$ is dense in $X$, $U\cap (\cup_n E_n) \neq \varnothing$, and there exists $n_0$ such that $U\cap E_{n_0} \neq \varnothing$.
Let $x \in U\cap E_{n_0} \neq \varnothing$.
Since $U$ is open, there exists $\varepsilon >0$ such that $B(x;\varepsilon) \subset U$.
Now, $B(x;\varepsilon)\cap E_{n_0}$ is open in $E_{n_0}$ by definition of the induced topology, and non-empty.
$F_{n_0}$ being dense in $E_{n_0}$, there exists $y\in F_{n_0}$ such that $y\in B(x;\varepsilon)\cap E_{n_0}$.
In particular, $y\in U\cap F_{n_0} \subset U\cap F$.
Therefore, $U\cap F \neq \varnothing$, and $F$ is dense in $X$.
