Explanation of the current-voltage relationship of a capacitor
How does this relationship proves that a discontinuous change in
  voltage requires an infinite current?


First, note that the equation given in your question defines an ideal (non-physical) capacitor and so this is the context of my answer.

Second, note that if the capacitor voltage is discontinuous at some instant(s) of time, the time derivative of the voltage does not exist there.

However, one can approximate a discontinuity by, e.g., letting the voltage change linearly with time over some short interval.  For example, let the capacitor voltage change linearly from \$0\mathrm V\$ to \$1\mathrm V\$ in \$\Delta t\$ seconds.

Then, according to the ideal capacitor equation, the capacitor current during the transition is

$$i(t) = C\frac{1 \mathrm V}{\Delta t}$$

In the limit as \$\Delta t \rightarrow 0\$, the capacitor voltage becomes discontinuous (finite change in zero time) and the capacitor current goes to an infinity large, infinitesimally short pulse; a current impulse.

But this is academic since physical capacitors obey the ideal capacitor equation only approximately and over a relatively narrow region of operation.