Yes the phase argument is a bit too simple. 

For starters, it would not explain anyons n 2D. Secondly, why should $\Psi(1,2)$ be *exactly* equal to $e^{2i\delta}\Psi(1,2)$ and not just proportional to it? Quantum mechanics is defined over a projective Hilbert space, so *any* phase factor would not affect the physical observable anyway.

The first thing you need to realise is that the permutation of the particles' labels is a *continuous* operation over a parameter space, which cannot reliably be captured by an instantaneous operator that just slaps on a phase factor:

[![enter image description here][1]][1]

If the particles return to the same original position, they trace out a closed loop $\gamma$ (about a point $a$) in this topological space $X$. The description of loops requires basic topology -- in particular, we can use the fundamental homotopy group $\pi_1$ which comprises all *topologically distinct* loops, that is the different ways one can permute the particles labels.

The smooth and continuous label exchange operation being a proper rotation in $d$ dimensions, it is represented by $SO(d)$.

So you just need to know the fundamental homotopy group of $SO(d)$:
* $\pi_1$ of $SO(2) = \mathbb{Z}$,
* $\pi_1$ of $SO(3) = \mathbb{Z}_2.$

So, exchanging the particles means tracing out a trajectory in parameter space. A phase is picked up:
$$ \Psi(1,2) \rightarrow \Psi(1,2) = e^{i\theta} \Psi(1,2),$$
which can be acquired in $\pi_1$ different ways:
* $2\pi / \mathbb{Z}_2$ in $d = 3$ (and $d>3$ for that matter), which means that the phase can be $2\pi / {1,2}$ that is $2\pi$ or $\pi$. Bosons and fermions;
* $2\pi / \mathbb{Z}$ in $d = 2$, hence anyons.

  [1]: https://i.sstatic.net/dOPaX.png