Is it safe to apply Kirchhoff's voltage law to a closed loop containing an inductance with unsteady current?

Yes, the Kirchhoff Voltage Law (KVL):

Sum of voltages (meaning potential drops) on all elements, connected to each other via perfect conducting wires in series into a closed circuit, is zero.

is valid for lumped element RLC circuits, so also for circuit with inductors. In real AC circuits designed not to radiate too much, voltage can be measured across any element and KVL can be validated experimentally. It is valid for common frequencies, up to hundreds of MHz and even higher to GHz range if parasitic elements are added to the model.

The whole theory of RLC circuits with harmonic voltage sources is derived from KVL being valid all the time, while currents and voltages change.

Some people say KVL is not valid for circuits with an inductor, since $\oint \mathbf E \cdot d \mathbf s \neq 0$ if inductor is in the circuit. However, that is actually not a problem for KVL, because KVL is formulated using potential drops, not integrals of total electric field. Potential drop across the inductor may be non-zero, even if total electric field in its coils is zero, because the drop is defined not by integral of total electric field, but by integral of the conservative component of that field (sum of the Coulomb fields of all charges).

In coils of an ideal inductor, net electric field is zero, so the conservative part of electric field exactly cancels the non-conservative part. This means potential drop on the inductor $V_L$ cancels the induced EMF in the inductor $\mathscr{E}_i$, so they have equal magnitudes, but opposite signs:

$$ \mathscr{E}_i = - L\frac{dI}{dt}, $$ $$ V_L = - \mathscr{E}_i = L\frac{dI}{dt}. $$

We can then the latter expression to write down circuit equations based on KVL.

In coils of a real inductor, which have non-zero resistance and also some capacitance, net electric field is non-zero, and thus potential drop does not cancel the induced EMF perfectly. Induced EMF is still $\mathscr{E}_i = - L\frac{dI}{dt}$, but potential drop can't be so easily expressed.

$\renewcommand{\vec}{\boldsymbol}$Yes, Kirchoff's law can be applied to inductive circuits.

Faraday's law of induction states that $$\oint \boldsymbol{E} \cdot \mathrm{d}\vec{l} = - \frac{\mathrm{d}\it\Phi_B}{\mathrm{d}t}\tag{1}$$ where $\vec{E}$ is the electric field and $\it\Phi_B$ is the magnetic flux passing through the loop at any given instant.

Now, the electric field generated in space is of two types, conservative and non-conservative. Therefore, $$\oint \vec{E} \cdot \mathrm{d}\vec{l} = \oint \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} + \oint \vec{E}_\text{non-conservative} \cdot \mathrm{d}\vec{l}.\tag{2}$$

The conservative field is produced by the accumulated charges in the circuit. Kirchhoff's second law states that the algebraic sum of all differences in potential around a complete circuit loop must be zero*, i.e. $$\oint \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} = 0 \tag{3}$$ which is to be expected by the definition of conservative fields. It states about the 'differences in potential'. We define a potential for conservative fields. The law does not make any statements about non-conservative fields. So, Kirchhoff's second law is correct and you can apply it to inductive circuits.

Applying the law to inductive circuits

Consider a closed circuit that has resistive wires of resistance $R$, a generator and an inductor of inductance $L$. Let us assume that the generator is creating a potential difference of $\mathcal{E}$ across its ends at some instant. At the same time, the potential differences across the resistor and the inductor are $iR$ and $ L\frac{\mathrm{d}i}{\mathrm{d}t}$.

How are these potential differences created? There are magnetic fields in the inductor and the generator. Consider the inductor; when the current passing through it is decreasing, an electric field is induced in it such that $\int \vec{E}_\text{non-conservative} \cdot \mathrm{d}\vec{l} = L\frac{\mathrm{d}i}{\mathrm{d}t}$ when integrated across the length of the inductor. (You can prove this using Faraday's law) Now, assuming that the charges in the circuit distribute themselves quickly, the charges distribute themselves across the ends of the inductor such that the non-conservative electric field is balanced by the electric field due to the accumulated charges. And the potential difference due to these accumulated charges is what you calculate when you apply Kirchhoff's second law to inductive circuits. Thus, $$\int \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} = L\frac{\mathrm{d}i}{\mathrm{d}t}$$ across the inductor. Similarly, charges get accumulated across the ends of the generator. And thus you can apply Kirchhoff's voltage law to a circuit containing inductors with unsteady currents.

*This statement is taken from Physics by Halliday, Resnick and Krane, 5th ed., Vol. 2. I cannot read German so I do not know what were the actual words used by Kirchhoff.

I did not include batteries in my discussion because the non-conservative forces in the battery which maintain the potential difference across the ends require quantum mechanics to explain them. Maxwell's equations alone cannot account for it. But still, $\oint \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} = 0$ is valid by the definition of conservative fields. In case of a battery, you can assume that some non-conservative (chemical in nature) forces maintain a constant potential difference across the ends by accumulating charges. See this.

Golden Rule: Don't use Kirchhoff's law. Use maxwell's equations. They are always correct.

Kirchhoff's law is just a special case of the Maxwell–Faraday equation. $$\nabla \times E = -\frac{\partial B}{\partial t}$$ or in a simpler way (strictly speaking, this is also a special case)

$$\oint E.dl = -\frac{d\phi_B}{dt}$$

The above formula can be interpreted as, if you go around a circuit and sum up the potentials ($E.dx = dV$), it must be equal to $-\frac{d\phi_B}{dt}$. In most circuits, the right hand side of the integral evaluates to zero. This is Kirchhoff's law.

Whenever you are in doubt, refer to the Maxwell's equations. They will always work.

Yes.

Voltage between two points is just a difference of potentials of these points. So, sum of voltages in in closed loop must always be zero, it's just mathematics:

$$(p_2 - p_1) + (p_3 - p_2) + ... + (p_1 - p_n) = 0$$

UPDATE.

If we have some electric circuit consisting of batteries, inductors, diodes, etc., the electric field is "potential". That means that it is possible to assign some number ("potential") to each point and the voltage between any two points would be the difference of above-mentioned potentials.

If the electric field is potential, the Kirchhoff's voltage law IS just a simple mathematical sequence of the fact that field is potential.

Electric field is NOT always potential. If there is a changing magnetic field, or some parts of an electric circuit are moving through magnetic field the electric field is not potential and you can not use KVL out of the box. Usually the effect of magnetic field can be substituted with some additional batteries and after that you are welcome.

Can you use KVL if there is an inductance in a circuit? Inductance coil is about changing magnetic field! But you can use KVL even in this case. The effect of magnetic field inside the coil is the same as an additional battery producing $L * \delta{I}/\delta{t}$, the effect of magnetic field on voltages outside the coil is much much smaller (approximately number-of-turns-of-the-coil times smaller).

How exactly you can use KVL in Alternating Current circuit? At any moment you can calculate potentials and the voltages on any element of the circuit. And the voltages at this (and any other) moment would obey KVL. But let's say peak voltages or root-mean-square voltages would not.