Proof that $ I(θ) = E[(s(θ;X))^2] $
Using the definition of Variance, as proved below:
$ \begin{align} \bar Var(Y) &= E[(Y-E(Y))^2]\\ &= E[Y^2-2YE[Y]+(E[Y])^2]\\ &= E[Y^2]-2(E[Y])^2+(E[Y])^2\\ &= E[Y^2] - (E[Y])^2 \end{align} $
Using the definition of Variance, as proved below:
$ \begin{align} \bar Var(Y) &= E[(Y-E(Y))^2]\\ &= E[Y^2-2YE[Y]+(E[Y])^2]\\ &= E[Y^2]-2(E[Y])^2+(E[Y])^2\\ &= E[Y^2] - (E[Y])^2 \end{align} $