Walter Lewin in his lectures says but all books and online resources go against him. Who is correct?
You probably misunderstood what Lewin is saying. I watched one of his lectures where he deals with inductor in a circuit and he gets the potential differences correctly, including the inductor. Here is the video of the lecture:
What he is criticizing is a different thing - that teachers and textbook authors explain Kirchhoff's law in circuits with inductors incorrectly: they assume
$$ \oint \mathbf E\cdot d\mathbf s = 0 $$
is valid even for circuit with inductor (it is not) and then rewrite this using partial integrals across the elements in the circuit, assuming (again incorrectly) that the integral over the wire of the inductor is $+LdI/dt$ (in fact, the integral is much lower and for ideal coil, zero since there is no field inside perfect conductor).
The Kirchhoff Voltage Law is really a practical rule to formulate circuit equations rather than a law of physics or a specific condition valid only in some cases. KVL is based on the fact that since potential is single-valued function of position, sum of drops of potential in a closed path is zero. KVL is not and does not derive from integral of total electric field being zero. Even if $\oint \mathbf E \cdot d\mathbf s \neq 0$, the sum of potential drops in a circuit is still 0. This is because electrostatic potential is a function of position, it does not depend on path, so one must get to the same value after a round-trip. In usual circuits, including low frequency circuits with inductors, difference of electrostatic potential is measurable and Kirchhoff's voltage law is valid. Drop of voltage across the inductor is $+L dI/dt$.
If there is a voltage across an inductor, then it's current will be changing. This is essentially the definition of what an inductor is.
If you're looking for a DC steady state solution, then there cannot be a voltage across the inductor. Because if there were, the current would be changing, which would mean the system is not at steady state.