Intuition first: with no internal resistance, what happens on the left side of the circuit is independent of what happens on the right side since they're in parallel.
With the internal resistance we have a voltage drop. The more current we ask to the battery, the more the drop. So if we turn on S, there will be more current, more dropout in the internal resistance. This means that bulbs A and B will shine a little bit darker since they have less voltage and current going through them.
Now let's apply some math:
Each lightbult is basically a resistance, let's call it R.
Right now the battery is feeding two bulbs in series. The resistance the battery is seeing is 2R.
If we add one bulb more (close S), which is in parallel, the equivalent resistance will be 2R in parallel with R, or 2R // R. This is equal to 2/3 R.
The total resistance seen by the battery has reduced. So the battery will have to provide a higher current. This means a bigger drop in the internal resistance.