When the external resistor increases in temperature, why is the calculated value for internal resistance higher?

Your logic is not flawed ( assuming the internal resistance is constant). However, if the internal resistance is much smaller than the load resistance, the increase in voltage across the load resistance may be difficult to measure. This is because a small change in the load resistance will have little effect on the cell’s terminal voltage. In other words the battery will approach being a constant (ideal) voltage source (terminal voltage = emf).

Hope this helps.

Since your emf is constant, less voltage must drop across the internal resistor

By constant emf, I understand that a voltimeter reads the same value between the poles of the cell, no matter if the external resistance is $10 Omega$ or more. So, by definition the voltage drop doesn't change, but the current decreases from its initial value when the temperature of the resistor increases.

Ohm's law states that if the external resistance goes up, the current goes down. The voltage is constant if you ignore the voltage source's internal resistance. If you account for the small internal resistance the voltage of the source goes a little up.

The flaw is in the statement 'Less voltage drop means less resistance.' A reduction in potential difference between two points can be caused either by a reduction in resistance or by a reduction in current, depending upon the circumstances.

The overall voltage drop across the internal and external resistance is shared across the two elements in proportion to the resistance they offer. If the external resistance increases, it will take a larger share of the overall voltage drop. In the extreme cases:

-where the external resistance is infinite, there will be no voltage drop across the internal resistance;

-where the external resistance tends to zero, all the voltage drop is across the internal resistance.