Why is Prewitt filter High Pass
That depends on the definition of high-pass filter. If you define a high-pass filter as a filter that has high response in the high frequencies in frequency domain, then the easiest way is to take a look at the magnitude of Fourier transform, (by definition). 

Applying Fourier transform (in Matlab)

 A = fftshift(abs(fft2(padarray([-1 -1 -1; 0 0 0; 1 1 1],[10 10]))));


results in the following image:



Now the interpretation - The part in the middle is the low frequencies. It has low response. There are 2 high responses, both with zero X frequency and some high Y frequency. That is not surprising since  we took a filter that detects edges in Y direction.



Why does it make sense to define high-pass filter in this way? 

Because a convolution can be thought as multiplication in Frequency domain. That is, if you have a signal $S$ and filter kernel $f$,

$F[S**f] = F[S] * F[f] $

Or, put in another way:

$ S**f  = F^{-1} [ F[S] * F[f] ] $

where convolution is denoted by $**$



There is another, more intuitive way, that does not involve Fourier transform. A response to a linear filter is strong when the underlying signal "looks" like the filter itself. Therefore, there should be a strong response to edges.