I am going though a past paper and in the mark scheme it states that when the switch is closed, the current through R (the resistor) increases, while in X it decreases. I understand that it decreases in X due to the "more paths" available for the charge carriers to travel and so they "split up", however I fail to understand how this affects the current through R.
I understand that it decreases in X due to the "more paths" available for the charge carriers to travel and so they "split up"
That's not why the current decreases in X; the current decreases in X because there is less voltage across X.
Why is there less voltage across X? Because there's more current through R.
Why is there more current through R? Because the current through R is the sum of the currents through X and Y.
Think about it this way. For simplicity, stipulate that X and Y are identical and further stipulate that the current through R does not change when the switch is closed. It follows that the current through X will be cut by half when the switch is closed (the other half is through Y).
But that would mean that the voltage across X (and hence Y since they are in parallel) would be less than before.
But if the voltage across X is less than before, the voltage across R is greater than before (the sum of the voltage across X and R must be the constant $V$) which, by Ohm's law, requires that the current through R has increased rather than remained constant as stipulated.
Thus, we cannot stipulate that the current through R does not change when the switch is closed without reaching a contradiction so it must be that the current through R does in fact increase when the switch is closed.
