Does it really "prove" Spin-statistics Theorem? In quantizing a scalar field, we impose commutation relation between the field operators by hand. On the other hand, anti-commutation relation is imposed between Dirac field operators by hand. As a consequence one gets Bose-statistics (two-particle wavefunction is symmetric) in the first case and the Fermi statistics (two-particle wavefunction is antisymmetric) in the second case. 
But does it really prove the spin-statistics theorem?
 A: I write below the statement of the aforementioned theorem which assumes, as hypotheses, the validity of so-called "Wightman axioms" in the four-dimensional Minkowski spacetime. 
You see that there is nothing imposed by hand. It is actually a no-go theorem. Quantizing free fields, it establishes in particular that the standard choice is the only possible.
Spin-Statistics theorem (Streater-Wightman's book Thm 4-10, adopting the signature +---):
For a general irreducible spinor field the "wrong" connection with statistics:
$$[\phi_a(x), \phi^\dagger_a(y)]_+ =0\quad \mbox{$\phi$ with integer spin}$$
$$[\phi_a(x), \phi^\dagger_a(y)]_- =0\quad \mbox{$\phi$ with half-odd integer spin}$$
and $(x-y)^2<0$, implies:
$$\phi_a(x)|vac\rangle =0\:.\quad (1)$$
In a field theory in which all fields either commute or anti commute, this also implies $\phi=\phi^\dagger=0$.
Identity (1) immediately implies that all n-point functions of the theory vanish so that the theory  turns out to be trivial. $|vac\rangle$ is the unique (up to phases) normalized vector state which is Poincaré invariant. 
