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So any state in the Fock space can be parametrized by \begin{equation} |\nu \rangle = | (i_1,n_1), \ldots, \rangle, \end{equation} where $n_\alpha$ is the number of particles that ocupy the state discribed by the quantum number (potentially more than one) $i_\alpha$. For fermions $n_\alpha = 0,1$. Let us now assume without loss of generality that $\alpha < \beta$. Let us also assume that $n_\alpha \neq 0$ and $n_\beta \neq 0$ or else the statement is trivial. Then \begin{equation} \begin{split} b_\alpha^\dagger b_\beta^\dagger |\nu \rangle &= (-1)^\sigma_\beta b_\alpha^\dagger |(i_1,n_1), \ldots , (i_\beta, n_\beta + 1), \ldots \rangle \\ &= (-1)^{\sigma_\alpha \sigma_\beta} |(i_1, n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots (i_\beta, n_\beta + 1), \ldots \rangle , \end{split} \end{equation} where $\sigma_\beta$ is the number of occupied states with $i < i_\beta$. This is exactly what you get from $\eta_\beta |\nu\rangle = (-1)^\sigma |\nu \rangle$ and likewise $\eta_\alpha |\nu\rangle = (-1)^\sigma |\nu \rangle$. Now if we let the operators act the other way around, \begin{equation} \begin{split} b_\beta^\dagger b_\alpha^\dagger |\nu \rangle &= (-1)^\sigma_\alpha b_\alpha^\dagger |(i_1,n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots \rangle \\ &= (-1)^{\sigma_\alpha \sigma_\beta + 1} |(i_1, n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots (i_\beta, n_\beta + 1), \ldots \rangle , \end{split} \end{equation} The additional minus sign comes from the fact, that there is now one more state with $i < i_\beta$ that is occupied, namely the $\alpha$-state. So this proofs $\lbrace b_\alpha^\dagger, b_\beta^\dagger \rbrace = 0$.