I am stumbling upon this question I asked myself. We know that if $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ are two sequences of a Hilbert space $(\mathcal{H},\langle\cdot,\cdot\rangle)$ such that $$ \begin{equation} x_n\rightharpoonup x \text{ and }y_n\to y \end{equation} $$ where $\rightharpoonup$ denotes the weak convergence, we have $$ \begin{equation} \langle x_n,y_n\rangle\to\langle x,y\rangle \end{equation} $$
Now, what about if the two sequences are weakly convergent, that is we have $$ \begin{equation} x_n\rightharpoonup x \text{ and }y_n\rightharpoonup y \end{equation} $$
Do we still have the result on the inner product convergence ?
We know an orthonormal basis weakly converges to $0$ because of Parseval equality: if $(e_n)_{n\in\mathbb{N}}$ denotes an orthonormal basis of $\mathcal{H}$, then for all $f\in\mathcal{H}$, we have $$ f = \sum_{n\in\mathbb{N}}\langle f,e_n\rangle e_n $$ with $$ \Vert f\Vert^2=\sum_{n\in\mathbb{N}}\vert\langle f,e_n\rangle\vert^2 $$ As the sum converges, the term $\vert\langle f,e_n\rangle\vert^2$ has to converge to $0$. As it is the case for all $f\in\mathcal{H}$, we have $e_n\rightharpoonup 0$.
Thus, we define $x_n=y_n=e_n$ for all $n\in\mathbb{N}$. We have $x=y=0$ but $\langle x_n,y_n\rangle=1$ for all $n\in\mathbb{N}$.