Dense subspace of $\ell_2$

For $2 \leqslant k \leqslant m$, let

$$f_{k,m} = \sum_{n = k}^{m} \frac{1}{\sqrt{n}\,\log n}\cdot e_n,$$

where $e_n$ is the sequence with $e_n(i) = 0$ if $i\neq n$ and $e_n(n) = 1$.

We have

$$\lVert f_{k,m}\rVert^2 \leqslant \sum_{n = k}^{\infty} \frac{1}{n(\log n)^2} \xrightarrow{k\to\infty} 0,$$

and

$$\lim_{m\to\infty} \sum_{n = k}^m \frac{1}{\sqrt{n}\,\log n}\cdot \frac{1}{\sqrt{n}} = +\infty.$$

Let $n \in \mathbb{N}\setminus \{0\}$. We will show that $e_n$ belongs to the closure of the given subspace. Thus let $\varepsilon > 0$. Choose $k > n$ such that $\lVert f_{k,m}\rVert < \varepsilon$ for all $m \geqslant k$. Then choose $m \geqslant k$ such that

$$s(k,m) := \sum_{r = k}^m \frac{1}{r\log r} > \frac{1}{\sqrt{n}}.$$

Then

$$e_n - \frac{1}{s(k,m)\sqrt{n}} f_{k,m}$$

belongs to the subspace, and

$$\biggl\lVert \frac{1}{s(k,m)\sqrt{n}}f_{k,m}\biggr\rVert < \frac{\varepsilon}{s(k,m)\sqrt{n}} < \varepsilon.$$

Since the subspace spanned by the $e_n$ is dense in $\ell_2$, the assertion follows.

Let us note that your set $V$ is actually a vector subspace. It is enough to show that $V$ separates points, i.e., if for all $f\in V$ and some $x\in \ell_2$ we have $\langle f,x\rangle = 0$ then $x=0$.

We argue by contraposition. Take $x\in \ell_2$ and suppose that $x\neq 0$. Then $x_k\neq 0$ for some $k$. Let $f$ be such that $f_k = 1$, for some $m\neq k$, $$f_{m}=-\frac{1}{\sqrt k}\cdot\sqrt{m+1}$$ such that $-\sqrt{m+1} \cdot x_m \neq 1$ and $f_n=0$ otherwise. This is possible as the sequence $(1/\sqrt{k+1})_{k=1}^\infty$ is not in $\ell_2$. Then $f\in V$ and $\langle f,x\rangle\neq 0$.