I'm reading a proof of some duality at page 33 of this lecture note. There is a paragraph
For each $\ell$ the sequence $\varphi_k^{(\ell)}-c_X$ is bounded in $L^{\infty}$ so $\overline{\left\{\varphi_k^{(\ell)}-c_X\right\}_{k \in \mathbb{N}}}$ is weakly compact in $L^p(\mu)$ for any $p \in(1, \infty)$ (for reflexive Banach spaces boundedness plus closure is equivalent to weak compactness). Let's choose $p=2$ then, after extracting a subsequence, we have that $\varphi_k^{(\ell)}-c_X \stackrel{L^2(\mu)}{\rightharpoonup} \varphi^{(\ell)}-c_X \in L^1(\mu)$ (since $L^2(\mu) \subset L^1(\mu)$ ) for some $\varphi^{(\ell)} \in L^1(\mu)$.
Here $\varphi_k^{(\ell)}, c_X \in L^p (\mu)$. A convex set is closed in weak topology IFF it is closed in norm topology. However, a set closed in norm topology is not necessarily closed in weak topology. It seems to me the bolded sentence is meant that
In a reflexive Banach space, if a set is bounded in norm and closed in weak topology then then it is compact and sequentially compact in weak topology.
Could you confirm if my understanding is correct?