Continuous voltage on IO Pins while 5V is not Connected? i have a small project where i just sucessfully destroyed my frist of 5 Digisparks (Attiny85).

In the Picture you see the Cirucit. I have 2 Liion cells connected in series to get high enough voltage to step it down to constant 5V. To these 5V some Neopixels and a Digispark controller are connected which will set the color for the Neopixels. A0 and A1 measure the Voltage of the Cells to signal the user if the Voltage is to low so it can manually be switched of.
While prototyping it i didnt have the 10k resistors in there.
Now what happend: While connecting the cells after each other the Controllers LEDs already lit up with the Frist cell. I guess it was powered trough the A0 pin (again i didnt have the 10k resistor in there),  so i guess the internal IO Protection diodes conductet the voltage trough and this way the controller got damanged. Since this happend it wont turn on Anymore.
The problem: since i have really small space i cant use a multiple pole switch to also remove the voltage from the IO pins and therefore i will always have voltage von A0 and A1 while there is no 5V Supplyvoltage. I would guess that the Series resistance on the pin would be enough to Limit the current trough them and this way protect the controller from damange. But i dont really have experience with that and hope that you guys can give me some hints what could also be a problem or what could help :) I really dont want to destroy a second one xD
Thanks in advance and have a nice day :)
For this I would recommend isolating the power from the batteries with P-channel MOSFETs and only switching them on when you want to sample the voltage. This means that there is no power consumption at all (aside from an absolutely minuscule leakage current) when turned off.
The whole setup would require 4 MOSFETs - two P-channel and two N-channel. They don't need to be big, since the currents involved are very small, but they do need to have &quot;logic level&quot; thresholds, ideally &lt; 1.5V (-1.5V for the P-channel).
The basic circuit looks like this:



simulate this circuit &ndash; Schematic created using CircuitLab
The circuit is made up of two identical segments - M1+M3 and M2+M4.  The operation is fairly straightforward:

M1 (M2) is kept in the OFF state by R1 (R2) pulling the gate up to source.
M3 (M4) can pull the gate of M1 (M2) LOW turning it on and allowing current to flow
Any GPIO pin (D7 in this example) can control both M3 and M4 together
R5 keeps M3 (M4) off when there is no power - otherwise it floats.
R3/R4 are your existing voltage divider.

The purpose of M3 and M4 is to both turn the GPIO (D7) into &quot;open drain&quot; so that it's either pulling down or floating high (rather than driving high at 5V) and to isolate the two pullup gate voltages. The same effect could be achieved by two diodes and one MOSFET but it's simpler to just use two MOSFETs and not have to get diodes as well.