Let $1<p<\infty$ and $x\in \ell^p$ with $\|x\|_p\le 1$. Then there exists a sequence $(x_n)$ with $\|x_n\|_p=1$ such that $x_n\to x$ weakly.
My attempt: By Riesz lemma, there exists a sequence $(z_n)$ such that $\|z_n-z_m\|_p\ge \frac12$ and $\|z_n-x\|\ge \frac12$ for all $m\ne n$. Moreover, as $\ell^p$ is reflexive, so $(z_n)$ has a weakly convergent subsequence, say $(z_{n_k})$.
I don't know how to proceed.
Edit: If $z$ is the weak limit of a $(z_{n_k})$, then the sequence $y_k=z_{n_{k+1}}-z_{n_{k}}+x$ converges weakly to $x$. But the normalized sequence $\left(\frac{y_k}{\|y_k\|}\right)$ might not converge weakly to $x$, right?