This is an attempt to respond positivally to the answer, hope there is no mistake...
$\forall A,B\subset \mathbb R^2$, $d(A,B):=\inf(\left\{||a-b||, a\in A, b\in B\right\})$
And $Conv(A)$ is the convex hull of $A$.
$Fr(A)$ the frontier of $A$.
Let $N>0$ be an integer. An **officer strategy** is any $x:\mathbb R^2\mapsto \mathbb (\mathbb R^2)^N$ , $v\mapsto (x_1(v),...,x_N(v))$ (continuous) fonction such that $\forall u,t\in \mathbb R^N$, $\mathcal D_N(u,v):=||x_1(u)-x_1(v)||+...+||x_K(u)-x_N(v)||\leq ||u-v||$.
> Claim : there exists $\delta\in \mathbb R_+$ such that $\forall t\in \mathbb R_+\forall i\in [N]:=\left\{1,...,N\right\}$, $\exists f: \mathbb R_+\to \mathbb R^2$ that is 1-lipschitzian, such that $||f(t)-x_i\circ f(t)||=:D(t)\geq D(0)\delta$
We fix $N$ and $x$ such that $\forall i\in [N], ||x_i(0)||\geq 1$ and pose $X_f=X=x\circ f= t\mapsto (X_1(t),...,X_N(t))$.
And also $X^* = \left\{X_1,...,X_N\right\}$ and $X^*(t)=\left\{X_1(t),...,X_N(t)\right\}$.
It is more convenient to use the continous paths than alternatives moves with laps etc... but one can easily be convinced that this statement implies the existence of a wining strategy for the fugitive with a $\delta$ that does not depend on the initial configuration.
As it was mentioned in comments, the fugitive (F) will be safe if he is out of $Conv(X^*)$ the convex hull of the set of officer, we can then procede gradually and show that, there is some $\delta'$ such that the fugitive can escape any convex hull of a choosen subset $ Y^*\subset X^*$, and then escape a bigger subset $Y^*\cup Y'^*\subset X^*$ (staying at no more then $\delta'^2$ from everyone) **with the condition that F will never return in** $conv(Y^*)$ anymore. Then if in $P$ steps of these kind that will take F gradually out
of $conv(X)$, we can deduce that he was able to escape with $\delta'^P$, hence $\delta'^ N$ can be took for our general $\delta$ (and even $\delta'^{(N/2)}$, because one can chose $Y$ to be of maximal cardinal (hense more than $N/2$). Then, it is sufficiant to prove that he can win the $Y$-game : i.e. :
> $\exists \delta'\in \mathbb R_+$, $\forall Y^*\subset X^*$, $0\notin Conv(Y)$, $\exists t'\in \mathbb R_+$, and $f:\mathbb R_+\to \mathbb R^2$, 1-lipsichian such that $f(0)=0$ and $\forall t\leq t',f(t)\notin Conv(Y^*(t)))$ and $\exists z\in X^*\setminus Y^*$, $f(t')\notin Conv(Y^*(t')\cup \left\{z(t')\right\})$
We will suppose wlog $Y$ maximal (for inclusion), i.e . that $f(0)=0\in \bigcap_{z\in X^*\setminus Y^*} Conv(Y^*\cup \left\{z\right\}):=S_Y$ eitherways we have nothing to do and can take $t'=0$. We want $f$ to escape $S_Y$ without even enter in $Conv(Y^*)$. This mean he has to cross a "tangent" to the convex hull of the set of the "back officers" $Y^*$, that passes through some "front officer" in $Z=X\setminus Y$. (the convex hull is not smouth so it is not a tangent, but I'll give precize definitions later). The two nearest "tangent" will be the "target" of the fugitives. They delimit (together with $Fr(conv(Y^*))$ the "cone like" $S_Y$ that the fugitive is **in** and that vertex $s_Y$ could be a target too, but there might be a real officer at this place... Note that the fugitive cannot be caught if this vertex is "officer free" and that every officer is into the symetric of the target cone, so it is quite intuitive that it should not be very difficult for the fugitive to touch these segments before getting cought, but we nead a bit more : to prove that he can escape at a distance of more than some $\delta'$ that does not depend on $X$.
Let's then set precizally the definitions that where anounced.
For all $a\in \mathbb R^2$, $A,B\subset \mathbb R^2$
$Tan_A(a)=Fr(Conv(A\cup \left\{a\right\}))\setminus Fr(Conv(A))$
$Tan_A(B)=\bigcup_{b\in B} Tan_A(b)$
There exists $s_A(B)\in \mathbb R^2$ such that $Conv(A\cup \left\{s_A(B)\right\})=\bigcap_{b\in B}Conv(A\cup \left\{b\right\}):=S_A(B)$ we call $target_A(B):=tan_A(s_A(B))=Fr(S_A(B))\setminus (Conv(A))$
By now I write $X,Y,Z,...$ instead of $X^*,Y^*,Z^*,...$ as soon as there is no danger of confusion. I will also write "$X$ at time $t$" instead of $X(t)$, etc...and if not precized it implies that it is "for some $t\in \mathbb R^2$", or $at the time that the situation occures".
Fugitive is going to follow the line $\lambda_0:= (0s(0))$ in the direction of $s(0):=s_{Y(0)}(Z(0))$ until he get to distance $1/4$ from $Z$. Of course if before that he crosses $target_Y(Z)$ at some point he can stop because he has win the "$Y$-game."
So let's suppose that he is exactly at a distance $1/4=d(f,Z)$ from,say $z_0\in Z$ the nearest officer $z_0$ (it is possible because $f$ and $X$ are continuous (lipschitzian) fonctions)
We distinguish two cases
1) At this time $t_0$, $s(t_0)= z_0(t_0)$
Then the $d_0:=d(Tan_Y(z_0),f)<1/4$, and as the angles defined by $Tan_Y(z_0)$ is always smaller than that of the semi-lines of $Tan_Y(f)$ the fugitive can go perpendiculary from $z_0$ at each time so that if he choses the right side (left/right) he get away from $z_0$ will without getting nearer $Y$ (that is at more than $1/2$ distance) while $d_0$ is decreasing , even worse if $Y$ is moving.
Neverthe less the speed of decreasion of $d_0$ might not be fast if $z_0$ is going "away" (eventually combining with $Y$ (hopeless**) moves but then at some point $f$ will some $Tan_Y(z_1)$ or cross $Tan_Y(z_0)$
**Indeed $Y$ is more far so their moves have less incidence on the "front" of the line they carry than the neighbours of the front according to Thales theorem, and nevertheless they give some advance to the fugitive.
2) $s(t_0)\ne z_0(t_0)$ same idea : he take the perpendicular to $(fz_0)$ to get closer to the tangant of $z_0$ that not make him nearer than $Y$. It is even worse for officers because $d(f,Target_Y)<d_0$ and one of $target_Y$ or $tan_Y(z_0)$ is moving less than half of the speed of $f$. Then we can take $\delta = 1/4$.
In conclusion we can repeat this operation not more than $N/2$ times (if we take $Y$ "maximal" not only for inclusion but according its cardinality)
Then we can take $\delta=2^{-N}$