Why doesn't electric potential decrease gradually across a wire? Let's assume that the resistance of a wire is zero. Now, suppose the wire has a length of 10 m and is connected to a battery with an emf of 10 V. According to my physics textbook, the electric field should be constant across the wire. Using the equation ΔV = -ʃE·ds, the voltage drop for a path Δs in the wire should be -EΔs. Also, the electric field should be of 1 V.m. Why then doesn't the electric potential decrease of 1 V for every meter in the wire? When solving problems with Kirchoff's laws, we assume that the voltage is constant until there is a resistor in the circuit and I don't understand why. Even if the resistance of the wire is zero, shouldn't the potential in the wire decrease gradually according to ΔV = -ʃE·ds?
 A: You are assuming the electric field within a perfect conductor in a steady state is non-zero. Let's go through what this would mean.
For starters if it really is a perfect conductor, it would mean charges would be accelerating within your conductor. In a steady state, this does not happen, for the current inside the circuit is constant.
It could also mean your charges are meeting some resistance, and thus doing work, causing a voltage drop, in which case, they would indeed need some electric field to push them past the resistance. But then, this means your conductor is not $0 \,\Omega$.
Thus, the electric field within the conductor not being zero implies either the conductor not being perfect, or charges being accelerated continuously to infinity. We can thus conclude the electric field within a perfect conductor is 0 in a steady state.
Thus, no, you would not get any sort of voltage drop across it. Current comes in, current comes out, and it has to fight nothing, because it's a perfect conductor, so no voltage drop across it, or across any segment of it.
A: 
If there is a battery across the wire, then there will be a voltage.

If the battery is physical, it has non-zero internal resistance, and then the emf will be dropped across this internal resistance when the zero resistance wire is connected across it, i.e., the battery terminal voltage will be zero. The resulting current through the zero resistance wire has a specific name: the short circuit current. See, for example, Battery Internal Resistance & Short Circuit Current
If the context is ideal circuit elements where the battery has zero internal resistance (is an ideal voltage source), then the KVL equation for the circuit you propose is invalid. For example, if the battery fixes the voltage across its terminals at $1.5V$, and the zero resistance wire (ideal short circuit) fixes the voltage across its ends to be $0V$, the resulting KVL equation is
$$1.5V = 0V$$
So, in the context of ideal circuit theory, this type of circuit is 'not allowed'.
