How to perform correct MCU interface analysis based on VIH,VIL,IIH,IIL?
I'm surprised that any 5V supply would burn out a 3.3V chip through 10k pull-ups. That would be less than 2mA through the input protection diodes. (I'm assuming the CC3200 has protection diodes -- it would be weird not to.) If the 3.3V supply can't sink current, maybe the voltage on its output capacitor could build up, but I'd expect the regulator to compensate for that. Perhaps the 2A supply behaves differently at power-up.

Regardless, the solution is the same -- don't put 5V on a 3.3V pin! As you can see from the datasheet, digital input currents are in the nanoamp range, so you should (almost) never have to worry about them. All you should have to do is:


Make sure the pull-up voltage is within the VIH range. Normally, you want the pull-up voltage to be the same as the IO voltage (or VDD if they're the same). If the current's low enough, you can use the same supply.
Make sure the pull-up supply can provide enough current when the pulled-up lines are driven low. In your case, that's:


$$I_{supply} = \frac {3.3 \mathrm V} {10 \mathrm{k\Omega} || 10 \mathrm{k\Omega} || 10 \mathrm{k\Omega} || 10 \mathrm{k\Omega}} = \frac {3.3 \mathrm V} {2.5 \mathrm {k\Omega}} = 1.32 \mathrm {mA}$$


Make sure the output pins (PORTx[0-3]) can sink enough current to pull their lines low. This is the \$I_{OL}\$ spec. According to the datasheet, \$I_{OL}\$ is configurable, and the lowest value is 2mA. Each IO only has to handle one pull-up, so:


$$I_{PU} = \frac {3.3 \mathrm V} {10 \mathrm {k\Omega}} = 0.33 \mathrm {mA}$$
$$I_{OL} &gt;= 2 \mathrm{mA}$$
$$I_{PU} &lt; I_{OL}$$