Why does more bandwidth guarantee high bit rate?
The simplest explanation is found in Shannon's equation:

$$C = B\log_2(1+S/N)$$

where C = channel capacity in bits/second
      B = channel bandwidth in Hertz
      S = signal power in watts
      N = noise power in watts

This equation relates the maximum channel capacity (C), that is the maximum data rate, as a function of channel bandwidth (B) and channel signal-to-noise ratio (S/N).  The bandwidth basically sets the limit on how many symbols per second can be sent. The signal-to-noise ratio, S/N, sets the limit on how many bits can be sent by each symbol.  If you consider the signal to be a square wave, it is clear that higher bandwidths allow higher frequency square waves to be transmitted.  Similarly, higher signal-to-noise ratios allow more bits for each symbol because more amplitude values can be discriminated at the receiver. You can increase data rate, without increasing bandwidth, by increasing transmitter power because that improves the signal-to-noise ratio which, by Shannon's equation, increases the channel capacity.  However, as the equation also shows, the ultimate channel capacity also depends on the bandwidth.  Thus, for the same transmitter power, the channel with the higher bandwidth will have the higher channel capacity.