Proving a subset is equal to the closure of a spanning set

It is straightforward to verify $\text{span}\{e_n|n\in\mathbb{N}\}\subseteq K$. Also, $K$ is closed (check that any convergent sequence in $l^2(\mathbb{N})$ taking values in $K$ converges to a point of $K$). Therefore, $\overline{\text{span}\{e_n|n\in\mathbb{N}\}}\subseteq K$. To see the reverse inclusion, let $(x_n)_n\in K$. We know that $\sum_{n}|x_n|^2<\infty$, so given any $\epsilon>0$, there is $N\in\mathbb{N}$ such that $\sum_{n>2N}|x_n|^2<\epsilon^2$. Then choose a sequence $(y_n)_n\in l^2(\mathbb{N})$ such that the first $2N$ terms of $(y_n)_n$ and $(x_n)_n$ are equal. For $n>2N$, set $y_n=0$. Then clearly $(y_n)_n\in \text{span}\{e_n|n\in\mathbb{N}\}$, and we have $\|(y_n)_n-(x_n)_n\|_2<\epsilon$. Thus, $(x_n)_n\in \overline{\text{span}\{e_n|n\in\mathbb{N}\}}$. We conclude that $K\subseteq\overline{\text{span}\{e_n|n\in\mathbb{N}\}}$.