Does the observed wave speed change if the observer is moving towards the source? We have the formula for the perceived frequency with a moving observer as follows:

Therefore, the perceived frequency does change for the observer.
Also, my textbook (Physics for the IB Diploma Coursebook by K.A. Tsokos) has this to say about the wavelength perceived by the observer.

Taken together, these two facts ($f \ne f'$ and $\lambda = \lambda '$) seem to imply that the perceived wave speed must be different to the actual wave speed (because $v = f\lambda$ and $v' = f'\lambda$). However, my high-school physics teacher says that the perceived wave speed and the actual wave speed are the same. Is my teacher correct in saying this?
 A: I assume that by "actual wave speed" you mean the speed of the wave in a frame of reference in which the source is at rest and by "perceived wave speed" you mean the speed as measured in the frame of the observer moving toward the source. Then yes, the perceived and actual speeds are different. This is consistent with the Galilean addition of velocities.
It's not clear what your teacher means by the two speeds being the same. Can you elaborate?
A: It is equivalent to you, standing still on the railway tracks, measuring the velocity of a train (sound wave) relative to the railway tracks (air) as $v_{\text{train relative to track}}$.
You then move along the railway track towards the train (sound wave) at a velocity relative to the railway tracks (air) of $v_{\text{you relative to track}}$.
You will measure the velocity of the train (sound wave) relative to you as   $v_{\text{train relative to you}} = v_{\text{train relative to track}}+v_{\text{you relative to track}}$
which is equivalent to
$v_{\text{sound wave relative to you}} = v_{\text{sound wave relative to air}}+v_{\text{you relative to air}}$
