Why can't we define a unique wavelength for a short wave train? 
Here we encounter a strange thing about waves; a very simple thing . . .namely, we cannot define a unique wavelength for a short wave train. Such a wave train does not have a definite wavelength; there is an indefiniteness in the wave number that is related to the finite length of the train. . .$^\text{1}$

Now, why can't we define a unique wavelength for a wave packet?
$^\text{1}$ Lectures on Physics by Feynman, Leighton , Sands. 
 A: I would say that an answer is that length of wave packet and width of spectrum are related by:
$\Delta\omega \Delta t\approx 1$
"Width of spectrum" here is characteristic range of frequencies that signal contains, that is width of Fourier transform of the signal. Infinite sine wave contains only 1 frequency, that is its spectrum/Fourier transform is infinitesimal thin delta function. Any other function will be constructed from many frequencies.
So that wave packet (wave train) in frequency space is defined by a bandwidth, rather than by delta function (like infinite sin wave).
See "wave packet" for more.
Simple example might demonstrate it. Delta function $\delta(t)$ in Fourier domain is constant, that is, contains all wavelengths/frequencies. Rectangular window turns into sinc function that occupies all spectrum as well.
Multiply any sine wave by rectangle (that is, create wave packet limited in time) and in Fourier space you will get convolution of delta function (defined by sine frequency) with sinc, which will give you a shifted sinc (with peak around sine function frequency).
