Yes, subsystems of an entangled state – if this subsystem is entangled with the rest – is always in a mixed state or "statistical mixture" which is used as a synonym in your discussion (or elsewhere).
If we're only interested in predictions for a subsystem $A$ in a system composed of $A,B$, then $A$ is described by a density matrix $\rho_A$ calculable by "tracing over" the indices of the Hilbert space for $B$:
$$\rho_A = {\rm Tr}_{i_b} \rho_{AB}$$
Note that if the whole system $AB$ is in a pure state,
$$\rho_{AB}= |\psi_{AB}\rangle\langle \psi_{AB}| $$
If $\psi_{AB}$ is an entangled i.e. not separable state, i.e. if it cannot be written as $|\psi_A\rangle\otimes |\psi_B\rangle$ for any states $|\psi_A\rangle$ and $|\psi_B\rangle$, then the tracing over has the effect of picking all the terms in $|\psi_{AB}\rangle$, forgetting about their dependence on the $B$ degrees of freedom, and writing their probabilities on the diagonal of $\rho_{AB}$. That's why the von Neumann entropy will be nonzero – the density matrix will be a diagonal one in a basis and there will be at least two entries that are neither $0$ nor $1$.
Take a system of two qubits. We have qubit $A$ and qubit $B$. There are 4 natural basis vectors for the two qubits, $|00\rangle$, $|01\rangle$, $|10\rangle$, and $|11\rangle$ where the first digit refers to the value of $A$ and the second digit to $B$. A general pure state is a superposition of these four states with four coefficients $\alpha_{AB}$ where $A,B$ are $0,1$, matched to the corresponding values.
If $\alpha_{AB}$ may be written as $\beta_A\gamma_B$ i.e. factorized in this way, the pure state is separable. $|01\rangle$ is separable, for example. If it is not, then it is entangled. For example, $|00\rangle+|11\rangle$ is not separable so it is entangled.
The mixed state is a more general state than a pure state. In this case, it is given by a $4\times 4$ Hermitian matrix $\rho$. The matrix entries are $\rho_{AB,A'B'}$ where the unprimed and primed indices refer to the values of qubits $AB$ in the bra and ket vectors, respectively. If these matrix entries may be factorized to
$$\rho_{AB,A'B'} = \alpha^*_{AB}\alpha_{A'B'}$$
for some coefficients $\alpha_{A'B'}$ and their complex conjugates that specify a pure state $|\psi_{AB}\rangle$, then the density matrix $\rho$ is equivalent to the pure state $|\psi_{AB}\rangle$ and we say that the system is in a pure state. In the more general case, $\rho$ can't be written as this factorized product but only as a sum of similar products. If you need at least two terms like that to write $\rho$, then the state is mixed and the von Neumann entropy is therefore nonzero.
Best Answer
That is not the right question to ask. The question should be: how does any particular observer describe system A. An observer who doesn't know about the entanglement (cut off from B) will describe A as a mixed state, while an observer who knows about the entanglement with B will describe them as entangled. As per conventional understanding, the whole universe should be in a pure state and the appearance of a mixed state anywhere means that the state is entangled with something we don't know about. So in essence, all mixed states arise because there is some correlation/entanglement/information an observer doesn't know about. There isn't any notion of a "real" mixed state.
No, it is not described by a definite state, or any pure state which is a linear combination of the basis states. It's described by a "mixed state", which is specified by its density matrix -- which tells you with what probability a projection gives $|0\rangle$ or $|1\rangle$. That's all the information you know about a mixed state: its density matrix.