Let $X$ be a normed space and $(x_n)$ is a Cauchy sequence in the norm sense. Also assume the $x_n \rightarrow x_0 $ weakly. Then $x_n \rightarrow x_0 $ in norm.
What I did:Take $ \varepsilon >0 $ since $x_n$ is cauchy there a $n_0$ such that $ |\!| x_n-x_m |\!|< \epsilon$ $\forall n,m \geq n_0$ From Hahn Banach there are $x^{*} _n\, \in X^{*}$ such that $|\!| x_n-x_0 |\!| = | x^{*}_n (x_n-x_0)|$ and $|\!| x^{*}_n |\!|=1$. Hence \begin{align} |\!| x_n-x_0 |\!| &= | x^{*}_n (x_n-x_0)|\\ &=| x^{*}_n (x_n-x_m+x_m-x_0)| \\ &\leq | x^{*}_n (x_n-x_m)|+| x^{*}_n (x_m-x_0)|\\ &\leq \varepsilon+| x^{*}_n (x_m-x_0)|. \end{align} Since the last inequality hold $\forall m \geq n_0$ we can take the limit in respect of $m$ and then we get $|\!| x_n-x_0 |\!| \leq \epsilon $ (Since $x_n \rightarrow x_0 $ weakly). And then by definition we are done. Where I saw this exercise there was a hint.
Hint Observe that $x_n \in x_m +\varepsilon B_X$ and $x_m+\varepsilon B_X$ is weakly closed. How do we proceed from there?