Why does measuring the spin of an entangled particle cause it to become unentangle?
"Entanglement" is a term describing economically the quantum mechanical state of a system of particles. It is a short hand way of saying : these particles are described by the solution of the Schrodinger equation, with a wave function that can predict the probability of finding the individual particles in a specific (x,y,z) each with specific quantum numbers". The conserved quantum numbers accompany the particles after the state interacts, and any measurement is an interaction.
A measurement is one instant in the accumulation of the probability function, of experimentally checking the theory. Many measurements will give the square of the wavefunction for this system of particles, which means we should prepare an ensemble of same condition particles.
When one measures the quantum mechanical condition of a particle , the original wavefunction is no longer valid, the particle is now in a new quantum mechanical state, no longer entangled in a coherent wavefunction with the others, as the boundary conditions are changed by the measurement of this one particle.
Quantum numbers that are conserved allow to know what are the possible collective quantum numbers of the remaining particles by the values carried by the particle measured.
The answer is simple: measurement causes the wave-functions to collapse.
It can be said that one of the fundamental properties that makes Quantum mechanics so strange is the idea of superposition, which is the property that if you have two physically valid descriptions of a state, then it is physically just as valid for a system to be in any linear combination of both states at the same time (think Schrödinger's cat).
Entanglement is just a particular example of a superposition of states. For instance you can describe the spin of two particles as both being up or both being down. Entanglement is a state where they are in a superposition of both spins being up and both spins down at the same time.
Now if you make a measurement of the spin of one or both of the particles, then this will cause your superposition to collapse into being either both up or both down thus destroying the special superposition state that is the entangled state.
What "entangled" means
You probably have put too much in your head for the meaning of the word "entangled." Let me fix that for you:
Two systems are entangled in quantum mechanics if the results of separate experiments on those two systems display strange correlations when you bring them back together and compare them.
Notice that not all correlations can necessarily be realized by QM. You see this a lot when you're looking at Bell's inequality violations: often the "maximum violation" allowed by QM in some situation is not the biggest that you could imagine if you could 100% play God with the measurement outcomes.
Notice also that there are not-strange correlations: for example, I have one ball, I drop it down a tube, it hits a divider and goes into one or another sealed opaque box without my knowing which one it's in. Now experiments like "do I hear a rattle when I shake the box?" are correlated: if I hear a rattle in this box, I don't hear it in that box. We could talk about entanglement here, but it is a "classical" entanglement.
A good example to keep in mind
To give an example of a "strange" correlation I often like to give an example of a 3-player game I call Betrayal. The idea is that there are 3 players on a team which endures a million "rounds" of experiments. In each round, we separate the 3 into completely isolated rooms, flash a command on a screen, and they have a few seconds to hit either a button labeled 1 or a button labeled 0. If they do "the right thing" as a team in all of the million rounds, they all get a billion dollars: so they're all working together as a team. The trick is that we sometimes try to force one person, who we'll call a "traitor", to work at cross-purposes to the other two. So there are two sorts of rounds. In control rounds we simply give them all the command, "make the sum of your team's chosen numbers even", and they all win the round if, when we take the three 1's or 0's that they press and sum them together, the result of this is even. In test rounds we choose a traitor at random and give them the same command as before -- make the sum even -- but we give the other two people the opposite command: "this is a traitor round, you make the sum of your team's chosen numbers odd." The team passes the round only if the sum of the 3 numbers they hit is odd.
You can prove with classical random variables that there is no 100% solution to this puzzle. (Basically you get $X_o + Y_o + Z_e \equiv X_o + Y_e + Z_o \equiv X_e + Y_o + Z_o \equiv 1 ~~\text{(mod 2)}$, sum those three equations together to get $$X_e + Y_e + Z_e + 2 (X_o + Y_o + Z_o) \equiv X_e + Y_e + Z_e \equiv 3 \equiv 1 ~~\text{(mod 2)},$$which contradicts the requirement that $X_e + Y_e + Z_e \equiv 0~~\text{(mod 2)}.$ So it does not matter what joint probability distribution you choose among these six random variables or how they correlate: classical probability cannot simultaneously give you all of these equivalences. Essentially the problem is that in classical probability we could ask a player $X$ for both answers $X_o, X_e$ and they could provide us with both values; it "doesn't matter" for a joint probability distribution that we didn't ask for the other one.
You can also easily prove with quantum mechanics that there is a 100% solution with three qubits in a GHZ state. Let the Hadamard basis be $|+\rangle = |0\rangle + |1\rangle$ and $|-\rangle = |0\rangle - |1\rangle$, then we have $$|+++\rangle = |000\rangle + |001\rangle + |010\rangle + |011\rangle + |100\rangle + |101\rangle + |110\rangle + |111\rangle$$ $$|---\rangle = |000\rangle - |001\rangle - |010\rangle + |011\rangle - |100\rangle + |101\rangle + |110\rangle - |111\rangle$$so that conveniently in the computational basis,$$|+++\rangle + |---\rangle = |000\rangle + |011\rangle + |101\rangle + |110\rangle$$ $$|+++\rangle - |---\rangle = |001\rangle + |010\rangle + |100\rangle + |111\rangle$$In other words, you all start off with the Hadamard-GHZ state $|+++\rangle + |---\rangle$ and anyone who is asked to make the sum odd simply performs a controlled-phase rotation in the Hadamard-basis: the unitary transform such that $|+\rangle\rightarrow|+\rangle, ~~ |-\rangle \rightarrow i |-\rangle$. If two people do this to their qubits in isolation, we flip from a state where all of the measurements have an even number of 1's to the state $|+++\rangle + i^2 |---\rangle = |+++\rangle - |---\rangle$, where all of the measurements have an odd number of 1's.
Why measurement destroys entanglement
There is a type-error in the question: a measurement doesn't destroy entanglement, it reveals it. You don't get to see entanglement unless you measure a bunch of things and bring the measurements together afterwards.
But measuring the wrong thing can destroy entanglement, sure. And that's for a familiar reason: in QM, once you measure X, often you cannot measure Y in the same way afterwards. You measure a spin up-down, you find it is up; then you measure it left-right, you find it is left; now you measure it up-down again: shockingly, sometimes it will be down now.
Similarly, if one of our hopeful candidates in the 3-member team accidentally measures in the Hadamard basis instead of the computational one, and they find their qubit is in the state $|+\rangle$, they have all lost the ability to do the measurements which display entanglement: the resulting system will just look like an unentangled $|+++\rangle$ system to all further experiments. One way to phrase this is that the entanglement was destroyed, but really what happened was that you performed a measurement which made it impossible for further measurements to reveal the entanglements in the system.
The Quantum Eraser Experiment
To give a great example of how this stuff can play out, let us consider a slightly different experiment: the quantum eraser. This is usually described in terms of a double-slit experiment where the states $|0\rangle$ and $|1\rangle$ are evolved into bell curves which overlap on our detectors, but the state $|+\rangle$ is evolved into an "interference pattern". The full treatment of what the pattern on the detectors looks like requires "state matrices" and "tracing out qubits" to really understand.
Quick overview: we start in the state $|00+\rangle$, we measure that third qubit in the Hadamard basis, we discover $+$, which is an "interference pattern", and we are happy. Actual details: the operator associated with this measurement is $$1 \otimes 1 \otimes |+\rangle\langle+| = |00+\rangle\langle00+| + |01+\rangle\langle01+| + |10+\rangle\langle10+| + |11+\rangle\langle11+|,$$and when we measure that (with appropriate factors of $\sqrt{1/2}$ inserted) we find an expected value of 1, which means that it's in the state $|+\rangle$.
Now we measure "which path does it go?" by performing a CNOT (quantum controlled-not gate) from qubit 3 to 2. This takes us to the state $|000\rangle + |011\rangle$. It turns out that the operator above expects only a value of $1/2$ in this state. This ultimately means "no interference pattern". We measured "which way" the photon went through the slits 0 and 1, and so we lose the glorious interference pattern due to it going through both with a quantum mixture $|+\rangle$ of 0 and 1.
But now suppose that the first two qubits were not starting in the state $|00\rangle$ but the state $|00\rangle + |11\rangle$. Then the first experiment is unaffected due to lack of entanglement, but the CNOT now puts us in the state: $$|000\rangle + |011\rangle + |110\rangle + |101\rangle = |+++\rangle + |---\rangle$$. Now we can again calculate the expectation value of the above operator and we see $1/2$: no interference pattern.
But: there is now a measurement we can make on a remote qubit which "erases" the which-way information: measure the first qubit in the Hadamard basis to find an explicit $|+\rangle$ or $|-\rangle$, and then you must see an interference pattern when you look at the screen. Interesting, no?