Let $1\le p<\infty$, and let $U$ be the open unit ball in $\ell^p$. Let $\{x_k\}\subseteq\overline U$ be a sequence such that for every $n$ the limit $x(n):=\lim_{k\to\infty}x_k(n)$ exists. Can I then conclude that $x\in\overline U$?
Now I've managed to show using the proof here that $x_n$ converges weakly to $x$.
Then theorem 3.12 in Rudin's Functional Analsyis tells me that the norm closure of $U$ is equal to the weak closure of $U$, so it follows that $x\in\overline U$.
But using that proof feels like overkill so I wonder if there's a simpler way.
A short proof is to apply Fatou's lemma to $\mathbb{N}$ with counting measure. It yields $$\sum_n|x(n)|^p \le \liminf_{k\to\infty} \sum_n|x_k(n)|^p \le 1$$ as desired.
For a self-contained proof one does what Daniel Fischer suggested. To prove $\|x\|_p\le 1$, it suffices to show that $\sum_{n=1}^N|x(n)|^p\le 1$ for every $N$; and the latter follows from coordinate-wise convergence.