dense subspace of $\ell_2$ from basics and seperability Why are the spaces spanned by $\{e_1-e_2,e_2-e_3,\ldots\}$ and $\{2e_1-e_2,2e_2-e_3,\ldots,\}$ dense but the spaces spanned by $\{e_2,e_3,\ldots\}$ and $\{e_1-2e_2,e_2-2e_3,\ldots\}$ not dense in $\ell_2$, where $e_i$ denotes the $i$-th coordinate vector, i.e. the sequence with 1 at the $i$-th place and zeros everywhere else?
How to prove the above facts from basics and from the fact the first two spaces mentioned separates points? I know that $\{e_i\}$ is an orthonormal basis of $\ell_2$ Any hints? Thanks beforehand.

Since $\ell^2$ is a Hilbert space, you can use that a subspace $M$ of $\ell^2$ will be dense if and only if $M^\perp = \{0\}$. This follows from Riesz projection theorem:$$\overline{M} \oplus M^\perp = \ell^2$$
Your first two subspaces indeed have trivial orthogonal complements.
Take $x \perp \operatorname{span} \{e_n - e_{n+1}\}_{n \in \mathbb{N}}$. For any $n \in \mathbb{N}$ we have $x \perp e_n - e_{n+1}$ so $$\langle x, e_n \rangle = \langle x, e_{n+1}\rangle$$
Hence $\langle x, e_n \rangle = \langle x, e_1 \rangle$ for all $n \in \mathbb{N}$.
Using Parseval's formula, we have:
$$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2 = \sum_{n=1}^\infty \left|\langle x, e_1\rangle\right|^2 \implies \langle x, e_1\rangle = 0$$
Hence $x = 0$ so the orthogonal complement is trivial.

Similarly, take $x \perp \operatorname{span} \{2e_n - e_{n+1}\}_{n \in \mathbb{N}}$. For any $n \in \mathbb{N}$ we have $x \perp 2e_n - e_{n+1}$ so $$2\langle x, e_n \rangle = \langle x, e_{n+1}\rangle$$
Hence inductively $\langle x, e_n \rangle = 2^n\langle x, e_1 \rangle$ for all $n \in \mathbb{N}$.
Using Parseval's formula, we have:
$$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2 = \sum_{n=1}^\infty 2^{2n}\left|\langle x, e_1\rangle\right|^2 \implies \langle x, e_1\rangle = 0$$
Hence $x = 0$ so the orthogonal complement is trivial.

On the other hand, you can similarly show that
$$\Big(\operatorname{span}\{e_n\}_{n\ge 2}\Big)^\perp = \operatorname{span} \{e_1\}$$
and $$\Big(\operatorname{span}\{e_n-2e_{n+1}\}_{n\in\mathbb{N}}\Big)^\perp = \operatorname{span} \left\{\sum_{n=1}^\infty \frac{1}{2^n}e_n\right\} = \operatorname{span} \left\{\left(1, \frac12, \frac14, \ldots \right)\right\}$$
so both subspaces have nontrivial orthogonal complements. Therefore, they are not dense in $\ell^2$.

Regarding point separation for the first subspace, you can again use that the orthogonal complement is trivial:
Assume that $$\langle x, e_{n} - e_{n+1}\rangle = \langle y, e_{n} - e_{n+1}\rangle, \quad \forall n \in \mathbb{N}$$
Therefore, $x - y\perp e_{n} - e_{n+1}$ for all $n \in \mathbb{N}$ so $$x - y \in \Big(\operatorname{span}\{e_n - e_{n+1}\}_{n \in \mathbb{N}}\Big)^\perp = \{0\} $$
We conclude $x = y$. Therefore, if $x \ne y$ then there exists $n \in \mathbb{N}$ such that $\langle x, e_{n} - e_{n+1}\rangle \ne \langle y, e_{n} - e_{n+1}\rangle$ so the first subspace separates points.
The second subspace is completely analogous.
