The measurement doesn't "destroy the whole quantum system", because you're not measuring the whole quantum system. You're only measuring two of the three systems.
It's like saying: if you have maximally entangled qubits, $|00\rangle+|11\rangle$ and you measure the first one in the computational basis, $\{|0\rangle,|1\rangle\}$, the second one is not "destroyed", but rather collapses on the same state you observed measuring the first qubit. If you measured $|0\rangle$, you know the second qubit is also $|0\rangle$, etc.
If you instead measure the first qubit in the $X$ eigenbasis, $\{|+\rangle,|-\rangle\}$, then finding $|+\rangle$ collapses the second qubit to $|+\rangle$, etc.
You can think of teleportation as working along the same principles. Except the measurement you're performing on the first qubit depends on the state $|\psi\rangle$ you're trying to teleport. The trick of doing a Bell measurement on two qubits on the sender's side achieves precisely this. Except instead of the sender performing a standard two-outcome projective measurement on their part of the entangled state, they perform a four-outcome POVM which depends on the state $|\psi\rangle$ to send over.
Formally, you can see it because the entire procedure on the sender side amounts to performing the POVM $\boldsymbol\mu$ such that:
$$\operatorname{tr}(\mu_i \rho)=\langle \Psi_i| \mathbb{P}_\psi\otimes \rho|\Psi_i\rangle, \qquad \mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|,$$
with $|\Psi_i\rangle$, $i=1,2,3,4$ the four Bell states, and $\rho$ the reduced state on the sender's side (yes, in the entanglement protocol this reduced state is just $I/2$, but here I'm describing the measurement procedure in isolation, I don't care what $\rho$ is; the measurement will then actually be performed on the entangled state, not actually just on the reduced state). Note that we can always write the Bell states as $|\Psi_i\rangle=(U_i\otimes I)|m\rangle$ with $|m\rangle\equiv\frac{|00\rangle+|11\rangle}{\sqrt2}$ and $U_i\in\{I,X,Y,Z\}$ the Pauli gates. It follows from some simple manipulation that
$$\mu_i = \operatorname{tr}_1[(\mathbb{P}_\psi\otimes I)\mathbb{P}_{\Psi_i}]
\equiv (\langle\psi|\otimes I)|\Psi_i\rangle\!\langle\Psi_i|(|\psi\rangle\otimes I)
= \frac12 \mathbb{P}(U_i^T|\bar\psi\rangle).$$
So what happens is: the sender gets the outcome $i$ from their four-outcome measurement, which collapses their state on $U_i^T|\bar\psi\rangle$. But projecting one part of $|00\rangle+|11\rangle$ on some $|\phi\rangle$ gives you $|\bar\phi\rangle$ on the other side, as follows from
$$(\langle\phi|\otimes I)(|00\rangle+|11\rangle) = |\bar\phi\rangle.$$
In other words, when the sender gets the $i$-th outcome, the receiver's state collapses to $\bar U_i|\psi\rangle$. Once they know which $i$ was measured, it is therefore easy for the receiver to correct away the $\bar U_i$ operation and end up with $|\psi\rangle$.
