Let $1\leq p<2$ and define $\ell^p:=\left\{(x_n)_n: \sum_{n\geq 1} |x_n|^p< \infty\right\}$. I want to check that for any $R>0$ $$B_R:=\left\{(x_n)_n: \sum_{n\geq 1} |x_n|^p\leq R\right\}$$is closed in $\ell^2$.
My idea was the following:
Proof Let me remark that $$\begin{align}B_R&:=\left\{(x_n)_n: \sum_{n\geq 1} |x_n|^p\leq R\right\} \\&=\left\{(x_n)_n:\left( \sum_{n\geq 1} |x_n|^p \right)^{1/p}\leq R^{1/p}\right\}\\&=\{x=(x_n)_n: ||x||_{\ell^p}\leq R^{1/p}\}=\bar{B}(x,R^{1/p})\end{align}$$ But we know that $\bar{B}(x,R^{1/p})$ is a closed ball in $\ell^p\subset \ell^2$ and hence it is also closed in $\ell^2$. Thus $B_R$ is closed in $\ell^2$ for all $R$
Does this work?
It works, but an objection would be that it is as hard to prove that $\bar{B}(x,R^{1/p})$ is a closed ball in $\ell^p\subset \ell^2$ as the initial problem (unless of course if it is supposed to be an application or if you are explicitly allowed to use it).
Here is a direct argument: $B_R$ can be expressed as $\bigcap_{N\geqslant 1}F_N$, where $$ F_N:=\left\{(x_n)_n\in\ell^2: \sum_{n= 1}^N |x_n|^p\leqslant R\right\}. $$ Therefore, it suffices to shows that for each $N$, $F_N$ is closed in $\ell^2$, which follows for example from the fact that for each fixed $N$, the map $(x_n)_n\in\ell^2\mapsto \sum_{n= 1}^N |x_n|^p$ is continuous.