How can the entropy of quantum states increase after projective measurements?

The key feature that you've left out of the question is the requirement that the measurement is performed, but we do not learn the answer.

For example, imagine starting with a $|+rangle$ state. You measure it in the $Z$ basis, so you get answers $|0rangle$ and $|1rangle$ with 50:50 probability. If you knew which result you'd got, then great, you have a pure state and 0 entropy. But if you don't know the result, your best description of the state is $frac12|0ranglelangle 0|+frac12|1ranglelangle 1|$ which obviously has 1 bit of entropy.

---

I might add that even if you do learn the measurement result, you are not guaranteed to have a lower entropy. For example, one of your projectors could project onto a maximally mixed subspace. Instead, it is only the average entropy that does not increase, which is perfectly compatible with some specific answers having higher entropy.

For example, let $$rho=(1-x)|00ranglelangle 00|+frac{x}{2}|1ranglelangle 1|otimes I.$$ $rho$ has eigenvalues $1-x,x/2,x/2$, so you can easily evaluate its entropy. Now, measure the first qubit in the $Z$ basis. With probability $1-x$, I get the answer $|0rangle$, the output state is a pure state, and hence has entropy 0. However, with probability $x$, we get the answer $|1rangle$, and the state of the second qubit is $I/2$, which has entropy 1. The average entropy is $x$, which is less than the original entropy of $x+h(x)$, where $h(x)$ is the binary entropy.