Weak convergence, together with convergence of norms, implies strong convergence in a Hilbert space.
Solution 1:
The result you want to show should be: if $x_n$ converges to $x$ weakly and $\lVert x_n\rVert\to \lVert x\rVert$, then there is convergence in norm. To see that, expand $\lVert x_n-x\rVert^2$ and use the fact that $\langle x_n,x\rangle\to \lVert x\rVert^2$.