can't that internal resistance be measured quite easily?
No, it cannot always be measured quite easily as it can be very small.
Now, if it's so small, why do we care?
Well, we normally use a 4-terminal measurement precisely when the resistance we're trying to measure is itself rather small.
If we have some uncertainty in the value of the probe resistance, than the subtraction you suggest would result in a large uncertainty in the extracted resistance value.
To be specific, consider a case where we're trying to measure a device with resistance R using a probe with resistance Rprobe.
We put a current I through the series combination of the device and probe and measure the voltage
V=I(R+Rprobe).
We then compute the resistance of the device as
R=(V/I)−Rprobe.
If there's an uncertainty δRprobe in the resistance of the probe, then the uncertainty in R is also δRprobe.
Therefore, the two-terminal measurement is bad in the case that the uncertainty in the probe resistance is an appreciable fraction of the resistance you're trying to measure.
In those cases, it's better to use a procedure that eliminates this uncertainty entirely.
There is a very common case where the uncertainty in the probe resistance is huge: contact resistance.
Suppose I'm trying to measure the resistance of some structure on a chip by touching that structure with probes connected to the readout electronics.
The contact resistance between the probes and the metal is not reproducible from trial to trial, so there's a big δRprobe.