How should two particles accelerate to have their separation Lorentz-contract? 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?
 A: 
Suppose we have two particles [...] How should they instead accelerate to have the distance Lorentz-contract?

A solution is arguably described in J. Franklin, "Rigid body motion in special relativity" (arxiv.org/abs/1105.3899). 
Using the notation from the question above: Given the trajectory, wrt. the (inertial) "laboratory frame", of "particle $2$" as
$$x_2[ t ] := \frac{c^2}{w} \left( \sqrt{1 + \frac{w^2~t^2}{c^2}} - 1\right) + x_2[ 0 ],$$
i.e. uniform acceleration of "particle $2$" with proper acceleration $ w $,
and provided that
$$0 \lt x_2[ 0 ] - x_1[ 0 ] \lt \frac{c^2}{w},$$
set the trajectory of "particle $1$" to
$$x_1[ t ] := \left( \frac{c^2}{w} + x_1[ 0 ] - x_2[ 0 ] \right) ~ \left( \sqrt{1 + \frac{w^2~t^2~c^2}{(c^2 + (x_1[ 0 ] - x_2[ 0 ]) ~w)^2}} - 1\right) + x_1[ 0 ],$$
i.e. uniform acceleration of "particle $1$" with proper acceleration $ w ~ \frac{c^2}{c^2 + (x_1[ 0 ] - x_2[ 0 ]) ~w}$.
Then it can be shown that a "family of momentarily co-moving pairs", $\{ (A_t, B_t) \}$, can be considered (or even be found) such that
(a) for any parameter value $t \ge 0$ the pair $(A_t, B_t)$ is at rest to each other, 
(b) with distances between any such pair equal to each other:
$AB_t = AB_0 := x_2[ 0 ] - x_1[ 0 ]$
(c) the speed of pair $(A_t, B_t)$ wrt. the (inertial) "laboratory frame" is $\frac{w~t}{ \sqrt{1 + \frac{w^2~t^2}{c^2}} },$ 
(d) particle $1$ met $A_t$ as it "turned around" wrt. pair $(A_t, B_t)$ (i.e. after "having approached $A_t$ from the direction of $B_t$", and before "going to further separate from  $A_t$ towards $B_t$"),
(e) particle $2$ met $B_t$ as it "turned around" wrt. pair $(A_t, B_t)$ (i.e. after "having approached $B_t$ from the direction opposite of $A_t$", and before "going to further separate from $B_t$ in the direction opposite of $A_t$"), and
(f) $A_t$'s indication of having met particle $1$ and $B_t$'s indication of having met particle $2$ were simultaneous to each other.
Since the distance between any pair of members of the (inertial) "laboratory frame" whose indications of "having been met and passed by $A_t$" and "having been met and passed by $B_t$", respectively, were simultaneous to each other is 
$$ \frac{x_2[ 0 ] - x_1[ 0 ]}{ \sqrt{1 + \frac{w^2~t^2}{c^2}} } = (x_2[ 0 ] - x_1[ 0 ]) ~ \sqrt{1 - \left( \frac{w~t}{ \sqrt{1 + \frac{w^2~t^2}{c^2}} } ~\frac{1}{c}\right)^2}, $$
where the right-hand-side is stated in terms of the speed value which appears in (c),
there applies the familiar "length-contraction" relation at least between the (inertial) "laboratory frame" and each "momentarily co-moving pair, $(A_t, B_t)$".
However, obviously, since with the described "solution" particle $1$ and particle $2$ are not at rest to each other, there is no value of "distance between them" defined to begin with. Indeed, not even the ping durations of either of these two particles with respect to other remain constant; i.e. unlike the two "ends of a train" which were described in this/my answer to the question about "an accelerating and shrinking train" (PSE/a/63110).
A: "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.
A: 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.
