I am trying to do the following exercise :
Let $X$ be a separable Banach space , M a bounded set of $X'$, the dual of $X$, show that for every sequence $f_n$ there is a subsequence that is weak* convergent to some $f \in X'$.
Now my tough process with this was that since $X$ is separable and the $f_n's$ are bonded since they are in $M$ we just need to see that $f_{n_k}(x)$ is cauchy for every $x$ in $A$, where $A$ is the countable set such that $cl(A)=X.$ Now this might have something to do with $A$ being countable but I dont know how to find the functions $f_{n_k}$ from $f_n$ such that we will have a cauchy sequence , does anyone have some advice? Thanks in advance!
Your approach is already a good idea.
Let me provide a sketch for the rest of the proof.
You can enumerate the $x\in A$ by $x_1,x_2,\dots$. An important first step would be to construct a subsequence $K_1\subset\mathbb N$ such that $\{f_k(x_1)\}_{k\in K_1}$ is Cauchy (here you need the boundedness of $M$). You could repeat this process and find a subsequence $K_2\subset K_1$ such that $\{f_k(x_2)\}_{k\in K_2}$ is Cauchy, and repeat this process countably many times.
Finally, pick a diagonal subsequence $K_\infty$. Then one can show that $\{f_k(x_j)\}_{k\in K_\infty}$ is Cauchy for all $j\in\mathbb N$.
Also, these things are strongly related to the Banach-Alaoglu theorem.