Is there a simple mathematical way to prove that measurement destroys entanglement?
I can see that this is indeed true if I just take a specific measurement on an entangled state. What I am looking for is a mathematical proof that this is true in general.
It is not true in general that measurement destroys entanglement. However, if the measurement basis consists of unentangled states and if the measurement outcome unambiguously identifies a specific basis element then this is indeed the case. Also, this type of measurement performed on a subsystem destroys its entanglement with other subsystems.
Projective measurement can be described by a basis $|\psi_k\rangle$ labelled by a set of outcomes $\lambda_k$ for $k=1,\dots,d$ where $d$ is the dimension of the underlying Hilbert space. If $\lambda_k$ are all distinct, then the measurement outcome unambiguously identifies a specific basis element. In this case, the observable describing the measurement is said to be non-degenerate. When such a measurement is performed on the state $|\phi\rangle$ then the outcome $\lambda_k$ occurs with the probability $|\langle\phi|\psi_k\rangle|^2$ and the post-measurement state is $|\psi_k\rangle$.
If the measurement is non-degenerate, then the post-measurement state is one of the elements of the basis describing the measurement. Consequently, such measurement destroys entanglement if and only if the basis state $|\psi_k\rangle$ corresponding to the observed outcome $\lambda_k$ is unentangled. In particular, non-degenerate measurement is guaranteed to destroy entanglement if all the basis states $|\psi_k\rangle$ are unentangled.
On the other hand, if the measured observable is degenerate, then the post-measurement state may preserve preexisting entanglement.
Even though the subsystems $A$ and $B$ of a composite system $AB$ in a pure state $|\psi_{AB}\rangle$ may be entangled with each other, the system $AB$ in a pure state is never entangled with a third system $C$. If the measurement is non-degenerate then, by arguments above, the post-measurement state is a pure state. Therefore, the non-degenerate measurement destroys entanglement of a system being measured with other systems. If the measurement is degenerate then it may preserve entanglement between the measured system and other systems.
In summary, measurement may destroy, preserve or create entanglement between subsystems that are jointly measured. If the measurement is non-degenerate then it destroys entanglement of the system with other systems not included in the measurement.
Suppose that three subsystems $ABC$ are in a state where $A$ is not entangled with the other subsystems and $B$ and $C$ are entangled with each other, e.g. $|0_A\rangle\otimes(|0_B0_C\rangle+|1_B1_C\rangle)/\sqrt{2}$. If we measure the subsystems $AB$ in the Bell basis, then we create entanglement between $A$ and $B$ and destroy the entanglement between $B$ and $C$. Indeed, the post-measurement state of $AB$ is a Bell state and the post-measurement state of $C$ is a computational basis state.
The question is not completely well-defined as stated, because the answer depends on what precisely you mean with "measurement", and what you assume the post-measurement states to be.
A possible interpretation is: consider a bipartite state $\rho$, and suppose some kind of measurement is performed on one part of it (and leaving the other untouched). The most general way to describe the process of performing some measurement, after which the state becomes something else conditionally to the measurement outcome, is via an entanglement breaking channel. For a proof that these always destroy entanglement, see this post.
