We know a result that if for each bounded linear functional $f$ from a normed space $X$ to $K$(its field), $f(x)=0$ then $x=0$. Now if $x_n$ in $X$ is weakly convergent to $x$ then for each bounded linear functional $f$ on $X$ , we have
lim$|f(x_n)-f(x)|=0$
$\implies$ lim$|f(x_n-x)|=0$
$\implies$$|$lim$f(x_n-x)|=0$ (Since modulus function is continous so we can take limit inside)
$\implies$$|f$$($lim $(x_n-x))|=0$ (Since $f$ is bounded so continuous and so we can take limit inside )
$\implies$ $f$$($lim $(x_n-x))=0$
$\implies$ lim $(x_n-x)=0$ (if for each bounded linear functional $f$, $f(x)=0$ then $x=0$ )
$\implies$lim $||(x_n-x)||=0$
This implies $x_n$ is strongly convergent to $x$, hence I proved weak convergrnnce implies strong convergrnnce. Where I am wrong because in general weak convergrnt does not imply strong convergrnt.