Cauchy in Norm and Weakly converge Implies Norm convergent 
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?
 A: Fix $\varepsilon>0$. Since $\{x_n:n\in\mathbb{N}\}$ is a Cauchy sequence we can find $N\in\mathbb{N}$ such that $n\geq m\geq N$ implies $x_n \in x_m+\varepsilon B_X$
Since $\{x_n:n\geq N\}\subset x_m+\varepsilon B_X$ and $x_m+\varepsilon B_X$ is weakly closed, then 
$$
x_0=w\lim\limits_{n\to\infty} x_n\in x_m+\varepsilon B_X.
$$ 
Thus $x_0\in x_m+\varepsilon B_X$, which can be reformulated as $\Vert x_m-x_0\Vert\leq \varepsilon$.
Finally for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $m\geq N$ implies $\Vert x_m-x_0\Vert\leq \varepsilon$, hence $$x_0=\lim\limits_{m\to\infty} x_m$$
A: For a fixed $\epsilon>0$, there is an $m$ so that for all $n\ge m$ we have $x_n\in x_m+\epsilon B_X$. Then we have $(x_j)_{j\ge m}\subset x_m+\epsilon B_X$; from this and the fact that $x_m+\epsilon B_X$ is weakly closed, we must have $x_0\in x_m+\epsilon B_X$. But then it follows that $\Vert x_n-x_0\Vert<\epsilon$ for all $n\ge m$.
A: We have that $x_0\in x_m+\varepsilon B_X$. Indeed, if it's not the case, $x_0$ is in a weakly open set, and we can find a neighborhood $V$ of $x_0$ for the weak topology such that $x_0\in V\subset \complement (x_m+\varepsilon B_x)$. But $x_n$ should be in this neighborhood for $n$ large enough. 
But in fact, both approach use Hahn-Banach (in the second it's used for the weak-closeness of the closed unit ball). 
