How to derive Pauli Exclusion Principle without assuming anti-symmetry?
The Pauli Exclusion principle is determined from experimental facts alone.
First, its consequences were observed and then it was incorporated in the theory. See Kaplan´s article for a more erudite answer. Also, see Towler´s exposition. Although, he dwells in foundational issues.

Important: they are not equivalent unless you consider only totally symmtric/antisymmetric functions! The antisymmetry is assumed, it cannot be derived rigorously from particle independence nor tested with experiments.

As for getting to the conclusion that the wavefunction of fermions has to be antisymmetric here is a weak argument without requiring the Pauli Exclusion principle.
You could also say that since you cannot follow the trajectory of particles therefore they are indistinguishable -- at least, if they are of the same kind. That is, the permutation of any two of them
$$ \Psi(\pmb x_1, \pmb x_2, ...) \to \Psi(\pmb x_2, \pmb x_1, ...) $$
maintains the same physical description
$$ |\Psi(\pmb x_1, \pmb x_2, ...)|^2 = |\Psi(\pmb x_2, \pmb x_1, ...)|^2.$$
The permutation can be described as the action of an operator that satisfies $P^2 = 1$.
There are only two ways to acomplish this. Either the wavefunction is symmetric (the eigenvalue of $P$ is $-1$) under the exchange or antisymmetric (eigenvalue $-1$).
$$ \Psi(\pmb x_1, \pmb x_2, ...) =  \Psi(\pmb x_2, \pmb x_1, ...) \quad \text{  Symmetric: bosons}$$
$$  \Psi(\pmb x_1, \pmb x_2, ...) = -\Psi(\pmb x_2, \pmb x_1, ...)\quad \text{Antisymmetric: fermions}$$

Therefore, if $\pmb x_1 = \pmb x_2$ then $\Psi(\pmb x_1, \pmb x_1, ...) = -\Psi(\pmb x_1, \pmb x_1, ...) = 0$. That is, Pauli Exclusion principle.

The reason as to why this argument, that is the one that was elaborated to justify Pauli Exclusion principle and is seen nowadays in most textbooks, is weak is neatly explained in the reference mentioned above.