question concerning weak star convergence. Given:
X seperable Banach space. X' its dual. We have $M\subset X'$ a closed unit ball in X'. Choose a sequence $(x_{n}$) of nonzero elements in X which is dense in X. Define 
$d(x',y'):=\sum\limits_{n=1}^n \frac{1}{2^n}  \mid{x'(x_n)-y'(y_n))}\mid $
Prove that $M\ni x'_m\rightharpoonup^*x'\in M$ iff $d(x_m',x')\to 0\:as\: m\to \infty $.
1.Do we need to use Banach-Alaoglu theorem? 
2.Is it somehow related with the Schur's property or may be it gives us something for the proof? I was reading  something which says that a Banach space has Schur's property if  weak convergence of a sequence implies convergence in norm? 
3.Or  may be Rosenthal's $
l_1$ theorem can play a role. Statement :"Every bounded sequence in a Banach space contains a subsequence that is either weakly Cauchy or equivalent to the unit vector basis of $l_1$."
Any help is appreciated.
 A: Let it be clear that items 2 and 3 have nothing to do with the simple (though important) fact stated in the first paragraph. You are to show that when $X$ is separable, the weak* topology of $X^*$ is metrizable on bounded subsets. 
Aside: this fact does come up in the context of Banach-Alaoglu, allowing for a nice proof without Tychonov's theorem (see sequential Banach-Alaoglu). But you don't need any big theorems to establish the fact itself. 
The proof consists of two parts. 
Part 1: $d(x_m',x')\to 0$ if and only if $x_m'(x_n)-x'(x_n)\to 0$ for every $n$. 
The direction $\Rightarrow$ is straightforward. For the direction $\Leftarrow$ split the series defining $d$ into "head" and "tail": the head is small by assumption, the tail is small because of $1/2^n$. 
Part 2 $x_m'\to x'$ in weak* topology if and only if $x_m'(x_n)-x'(x_n)\to 0$ for every $n$. 
Again,  $\Rightarrow$ is straightforward. For the converse, fix $x\in X$. You can find $x_n$ close to $x$. Use the triangle inequality like this: 
$$\begin{split}  |x_m'(x_n)-x'(x)| &\le  |x_m'(x_n)-x'(x_n)| + |x_m'(x_n)-x_m'(x)| + |x'(x_n)-x'(x)|
\\ &\le |x_m'(x_n)-x'(x_n)|+2\|x-x_n\| \end{split}$$
