Question: Suppose we have two particles with initial separation of $|x\_2^\text{init}-x\_1^\text{init}|=d$. At time $t=0$ they start accelerating with proper acceleration of $w$. Then position of the $i$th particle in laboratory frame will be:

$$x\_i(t)=\frac{c^2}w\left(\sqrt{1+\frac{w^2t^2}{c^2}}-1\right)+x\_i^\text{init}.$$

But then we can see that
$$x\_2(t)-x\_1(t)=d=\text{const}.$$

So, if both particles uniformly accelerate starting from some equal moments of time, distance between them doesn't Lorentz-contract. Why is it so? How should they instead accelerate to have the distance Lorentz-contract? Should they continuously synchronize their velocities according to *laboratory* frame times instead of their proper times?

Answer: 
"Length contraction", is the measure of a umproper length of an object, in a inertial frame $O'$ which has a relative velocity to a inertial frame $O$, where the coordinates of the object are fixed in this frame $O$ 

More precisely, one measures a proper time $\Delta \tau'$ in $O'$ (so at $x'=0$), which corresponds to the synchronisation with the beginning and the end of the object, and this proper time is traduced into an unproper length : $L' = v \Delta \tau'$. Lorentz transformation implies $c^2\Delta \tau' = c^2 (\dfrac{L}{v})^2 - L^2 $. The result is then a "length contration" : $L' = \dfrac{L}{\gamma(v)} =\sqrt{1-\dfrac{v^2}{c^2}} L$

So, first, unless you modify your question, you are not in this situation, because you have not $2$ different inertial frames moving with a constant speed relatively to each other. Secondly, as indicated above, what is measuring is not directly a length, but a proper time in the moving (relatively to the object) frame. This proper time is traduced into a unproper length as indicated above.
To make the example (and the question) clear, the particles should be on different frames, where when the 2 frames are at rest and coincide, the particles have the given separation.

Then when **one frame** moves relatively (or *accelerates*) **to the other** with a given velocity (what velocity exactly?) the Lorentz-contraction appears **but only with respect to the frames each particle is in**.