Let $M \subseteq B(H)$ be a von Neumann algebra. Denote $B(H)_*$ to be the $\sigma$-weakly continuous functionals on $B(H)$ and let $M_*= \{\omega\vert_M: \omega \in B(H)_*\}$. I want to prove that there is an isometric isomorphism $M \cong (M_*)^*$. The book I'm reading just says that "this follows by general Banach space theory".
My main problem is that I don't see an obvious map $$M \to (M_*)^*$$ Given $m \in M$, we want to build a map $\psi_m: M_* \to \mathbb{C}$. Maybe $$\psi_m: M_*\to \mathbb{C}: \sigma \mapsto \sigma (m)$$
works? In other words, is the canonical isomorphism $M \cong (M_*)^*$ given by $$m \mapsto ev_m$$?
But you do see an obvious map. You wrote it in your question.
Let $\Gamma:M\to(M_*)^*$ be $\Gamma(x)=\hat x$, where $\hat x(\varphi)=\varphi(x)$. This is clearly linear. It is also isometric, because the norm can be obtained as $\|x\|=\lim_n\langle x\xi_n,\eta_n\rangle=\lim \hat x(\varphi_{\xi_n,\eta_n})$ for appropriate unit vectors.
So one needs to check that $\Gamma$ is onto. Without writing many details, what happens is this. The key fact is to notice that normal functionals on $M$ are restrictions of normal functionals on $B(H)$, and that these are given by the trace-class operators. That is, $$ M_*=B(H)_*/M^\perp, $$ where $B(H)_*$ are the trace-class operators with the usual trace duality, and $$ M^\perp=\{\varphi\in B(H)_*:\ \varphi|_M=0\}. $$ Then $$ (M_*)^*=(B(H)_*/M^\perp)^*\simeq M^{\perp\perp}=\overline{M}^{w^*}=M. $$