Finding a finite set of functions in a weak topology space

Question

Let $M(\Sigma)$ be the set of finite and signed measures on the Polish space $\Sigma$ and $M(\Sigma)^*$ be it's dual. We endow $M(\Sigma)$ with the weak topology. Then, by the duality relation, $$(\phi, \alpha) \in C_b(\Sigma;\mathbb{R}) \times M(\Sigma) \mapsto \int_{\Sigma} \phi d\alpha$$ one can identify $M(\Sigma)^*$ with $C_b(\Sigma;\mathbb{R})$, the set of bounded and continuous functions on $\Sigma$. I am trying to prove this statement. First, I want to show that every element of $M(\Sigma)^*$ arises in the following way: $\forall \phi \in C_b(\Sigma;\mathbb{R}), \alpha \in M(\Sigma) \mapsto \int_{\Sigma} \phi d\alpha$. For that aim, let $\lambda \in M(\Sigma)^*$ be an arbitrary element of the dual. We set $\phi(\sigma) = \lambda(\delta_{\sigma}), \sigma \in \Sigma.$ Then,$ \int_{\Sigma} \phi (\sigma')d(\delta_{\sigma}) = \sum_{\sigma' \in \Sigma}\phi(\sigma')d\delta_{\sigma}(\sigma') = \phi(\sigma)$. I am not sure about what follows: Given that we endow $M(\Sigma)$ with the weak topology (which by the way has a countable basis), we can find a finite set $(\psi_m)_{m=1}^M \in C_b(\Sigma;\mathbb{R})$ such that $$|\lambda(\alpha)|\leq\sum_1^M |\int_{\Sigma} \psi_md\alpha|, \alpha \in M(\Sigma).$$ We need finitely many functions in order to make $\phi$ be bounded. All the measures are finite, so the integrals are finite. Can somebody provide some support in proving the inequality based on the underlying (weak) topology? Thanks.