Existence of a subsequence such that $\lim_{n\to \infty} \|x_{n_k}-x\|=\ell>0$ in a Banach space? Let $x_n$ be a sequence in a uniformly convex Banach space $E$ such that $x_n \to x$ weakly in $E$ and $\|x_n\|_E \to \|x\|_E$. Then $x_n \to x$ strongly in $E$.
To show this by contradiction a proof I'm reading states that first we suppose that (for the non-trivial $x\neq 0$ case)
$$
\limsup_{n\to \infty} \|x_n-x\|>0.
$$
Then there exists a subsequence $x_{n_k}$ such that
$$
\lim_{n\to \infty} \|x_{n_k}-x\|=\ell>0.
$$
My question is how can we say there exists such a subsequence that converges to a finite number? I don't see anything in the assumptions that gives us this?

The point at which you seem to have a question is not about Banach spaces; it's about sequences of real numbers. You have a sequence $\{a_n\}_{n=1}^\infty$ of real numbers for which
$$
\limsup_{n\to\infty} a_n >0
$$
and the question is: how does this imply that there is a subsequence $\{a_{n_k}\}_{k=1}^\infty$ for which
$$
\lim_{k\to\infty} a_{n_k} = \ell>0.
$$
If you have $\limsup_{n\to\infty} a_n = \ell>0,$ then how do you show there is a subsequence converging to $\ell\text{?}$
First approach:
Apply the definition of $\limsup:$
$$
\limsup_{n\to\infty} a_n = \inf\left\{ \sup\{a_n,a_{n+1},a_{n+2}, a_{n+3}, \ldots\} : n\in\{1,2,3,\ldots\} \right\} = \ell >0.
$$
This means every number $h<\ell$ fails to be a lower bound of the sequence
$$
\Big\{ \sup\{a_n,a_{n+1},a_{n+2}, \ldots\} \Big\}_{n=1}^\infty.
$$
Failure of $h$ to be a lower bound of this sequence means some member of this sequence is $>h.$ Thus we have
$$
\forall h<\ell\  \exists n\  \sup\{a_n,a_{n+1},a_{n+2},\ldots\} > h.
$$
Just let $n_k$ be some index $\ge n$ for which $a_{n_k}>h.$
Second approach: Relying on a definition of $\limsup,$ show that $\limsup_{n\to\infty} a_n = \ell$ if and only if $\ell$ is the largest of all limits of subsequences of $\{a_n\}_{n=1}^\infty.$ That implies there is some subsequence converging to $\ell.$
