Why is the potential drop zero across a wire of zero resistance? Okay, even if the wire has zero resistance, then charges are being transferred in the direction of electric field, which means there must be a potential drop according to the distance travelled by the charges in the direction of electric field. But, in my book, as it is explained, it seems like the potential remains constant throughout the journey of the charge and that it abruptly changes when the charge reaches the other end of the battery.
Well, that's just like saying that a body dropped from a height has an abrupt potential drop when it reaches the ground and has no potential drop along its journey if there is no air friction.
I think the potential drop still happens. The difference is only that in case of an ideal wire, the potential drop corresponds to the increase in kinetic energy of the electrons, because the electrons can accelerate forever in this case when there are no positive ions to collide to, while in case of a real wire, the potential drop mostly corresponds to the heat developed.
 A: Zero resistance for a conducting wire is an approximation, one that's so close to reality and so useful that it often gets treated as fact.  Here's a table that lists AWG 22 copper wire (which is the smallest hole on my wire stripper, diameter 0.6 mm) as having a resistance of 53 milli-ohms per meter.  So imagine a circuit like this:
      -------------------- one meter thin wire -------------------------
 1 V battery                                                       100 ohms
      -------------------- one meter thin wire -------------------------

If your "thin wires" behave like the thin wires in my table, then each of them is a "resistor" with roughly $0.050\,\Omega$ resistance, and the voltage drop across the resistor is closer to $0.999\rm\,V$ than to the $1.000\rm\,V$ across the battery.  But most electronic components have manufacturing tolerances of about 1%, so that'd be a tricky difference to measure. It's safe to leave out when you're teaching circuits the first time.
Finite resistance in a long cable becomes important if you're sending a lot of current a long way, but for analyzing a circuit that fits in your hand you can usually neglect the transmission line.
A: Apply Ohm's law $\Delta V = iR$ across the two points, you'll see, since $R = 0\implies\Delta V = 0$.
Explanation:
Consider the image:

The wire is of zero resistance. $EMF$ of battery is $E$ volts.
From ohm's law, current $I$ in the circuit is $I = \dfrac{E}{R}$.
Now you may apply $KVL$ to find potentials at A, B, C. When you calculate, you'll find out that $$V_A = V_c$$
This result shouldn't be surprising even though electrons have externally flowed from $C$ to $A$, because Electric field exists only inside the resistor! 
That electric field has given the electrons some kinetic energy at point $C$, and they flow with the same velocity. There is no electric field inside the zero resistance wire.
Here is a nice visual example from this post on Physics.SE:

A: Another way to look at it is if you connect across a potential with a zero ohm wire the potential difference gets reduced to zero.  It would be impossible to maintain a potential under zero ohms.
