Subspaces of $\ell^{2}$ and $\ell^{\infty}$ which are not closed?

Are there any examples of subspaces of $\ell^{2}$ and $\ell^{\infty}$ which are not closed?

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Take the subspace of $\ell^2$ (or $\ell^\infty$) consisting of sequences with finitely many non-zero coordinates.

Clearly, this is a subspace (sum of sequences still has at most # of nonzeros of first seq + # of nonzeros in second seq, scaling of a sequence has same non-zeros). However, you can take any sequence on $\ell^2$ (or $c_0 \subset \ell^\infty$, a closed subspace), $a$ and define the sequence of sequences $\{a_n\}$ where $a_n$ consists of only the first $n$ elements of $a$ and rest zeros, and see $a_n \to a$.

Yes. For instance, the sequences which are eventually $0$.

As an observation, every subspace generated by countably many linearly independent vectors (that is, a subspace of dimension $\aleph_0$) inside a Banach space is not closed, since a Banach space cannot have (algebraic) dimension $\aleph_0$. ( if it were closed, if would be itself a Banach space).

For $1< x<2$ the space $l^x$ is a non-closed vector subspace of $l^2.$

Let $(A_j)_{j\in \mathbb N}$ be a strictly increasing positive real sequence converging to $1/x.$

For $i,j, \in \mathbb N$ let $B_{i,j}=0$ if $j>i,$ and $B_{i,j}=j^{-A_j}$ if $j\leq i.$ Let $v_i=(B_{i,j})_{j\in \mathbb N}.$

Since only finitely many co-ordinates of each $v_i$ are non-zero, we have $\{v_i:i\in \mathbb N\}\subset l^x.$

Let $u= (u_j)_{j\in \mathbb N}=(\;j^{-B_{j,j}}\;)_{j\in \mathbb R}=(\;j^{-A_j})_{j\in \mathbb N}.$

Take $d\in (0,1-x/2).$ Take $j_0$ such that $j\geq j_0\implies A_j>(1-d)/x.$ Then $j\geq j_0 \implies 2A_j>2(1-d)/x>1.$ So $$\sum_{j=j_0}^{\infty} u_j^2=\sum_{j=j_0}^{\infty}j^{-2A_j}\leq \sum_{j=j_0}^{\infty}j^{-2(1-d)/x}<\infty.$$ Therefore $u\in l^2.$

Now $u\not \in l^x$ because $u_j^x=j^{-xA_j}>j^{-1}$ so $$\sum_{j=1}^{\infty} u_j^x\geq \sum_{j=1}^{\infty} j^{-1}=\infty.$$

I will leave it to you to show that $\lim_{j\to \infty} \|u-v_j\|_2=0.$ So in $l^2,$ the vector $u$ belongs to $Cl(l^x)$ but not to $l^x.$

The fact that $l^x$ is a vector space for any $x>1$ is another story, which I am sure you can find on this site.