If the particles never did interact in the past and won't ever interact in the future, then it indeed makes not much sense to model the system as a tensor product. Keep in mind that it does not matter how long ago the interaction happened as long as the particles are not otherwise disturbed - that's one of the points of the EPR 'paradoxon'.
However, care must be taken when dealing with identical particles: They are excitations of the same underlying field and should always be considered as having interacted.
In fact, the state space of identical particles is not the tensor product, but its symmetrization (in case of bosons) or anti-symmetrization (in case of fermions), ie only entangled states are considered.
These are just my thoughts and by no means authorative - anyone who deals with such issues on a more hands-on basis probably has a better grasp of the underlying issuess...
Considering the unexplained downvote, a more technical answer to the problem:
In non-relativistic Quantum mechanics, the evolution of a system in governed by inital state and Schrödinger's equation.
If we start out in an un-entagled state and the Hamiltonian does not contain interactions between the different sub-systems, we will stay in an unentangled state and we can consider the sytems independantly.
More formally, let our inital state be
$$\psi = \psi_1\otimes...\otimes\psi_n$$
and the Hamiltonian given by
$$H = \tilde H_1 + ... + \tilde H_n$$
$$= H_1\otimes1\otimes...\otimes1 + 1\otimes H_2\otimes1\otimes....\otimes1 + ... + 1\otimes...\otimes1\otimes H_n$$
Then the state after time $\Delta t$ is given by
$$
\mathrm{e}^{-iH\Delta t}\psi = \mathrm{e}^{-i\tilde H_1\Delta t}\cdot...\cdot\mathrm{e}^{-i\tilde H_n\Delta t}\psi_1\otimes...\otimes\psi_n = (\mathrm{e}^{-iH_1\Delta t}\psi_1)\otimes...\otimes(\mathrm{e}^{-iH_n\Delta t}\psi_n)
$$
where the first equality follows from $[\tilde H_i,\tilde H_j] = 0$.
The canonical commutation relations are not well-defined on finite-dimensional Hilbert spaces. The canonical prescription is
$$ [x,p] = \mathrm{i}\hbar\mathbf{1}$$
and, recalling that the trace of a commutator must vanish, but the trace of the identity is the dimension of the space if it is finite-dimensional, we conclude that we have a space for which the trace of the identity is not well-defined, which is then necessarily infinite-dimensional.
Best Answer
It doesn't, but that's not what Scrinzi is saying.
The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation: $$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle x+x'|\rho|x-x'\rangle e^{-2ipx'/\hbar}\,\mathrm{d}x'\text{,}$$ where for pure states $\rho = |\psi\rangle\langle \psi|$ as usual, and Hermitian operators correspond to functions through the inverse Weyl transformation.
Note that $W(x,p)$ is a real function that's just like a joint probability distribution over the phase space, except that it's allowed to be negative. The uncertainty principle requires us to give up something, but it doesn't actually force Hilbert spaces on us.
However, what Scrinzi is saying is two things: (a) Hilbert spaces are very convenient to us in quantum mechanics, and (b) Hilbert spaces could be used in classical mechanics, but because non-commutativity doesn't exist in classical mechanics, it's "overkill" there, whereas it's "just the right amount of kill" in quantum mechanics. Both claims are correct.
The reason we can have used Hilbert spaces in classical mechanics is because they can represent very general algebras of observables, while the classical algebra of observables, being commutative, is actually simpler. (Cf. the Gel'fand–Naimark theorem for $C^*$-algebras in particular.)
The Hilbert space formulation of classical mechanics was done by Koopman and von Neumann in 1931-1932. But what their formulation actually does is generalize classical mechanics unless one imposes an artificial restriction that you're only ever allowed to measure observables in some mutually commutative set; only then is the classical (in the 19th century sense) mechanics is recovered exactly.
It is that artificial restriction that quantum mechanics lifts. Physically, non-commutativity observables corresponds to an uncertainty principle between them.