Interpret the set $S$ geometrically

The condition is $|z-2+i|\ge \sqrt{5}.$

Write $z-2+i = z-(2-i).$ Hence $S$ is the set of all points in the complex plane whose distance from the point $2-i$ is at least $\sqrt5$.

So $S$ is the exterior (including boundary) of the circle with center $C=(2,-1)$ and radius $r=\sqrt5.$

Use the maximization condition

We are told that $\frac1{|z_0-1|}$ is maximum among all such values for $z\in S$.

Maximizing $\dfrac1{|z-1|}$ is equivalent to minimizing $|z-1|$.

So $z_0$ is the point in $S$ closest to the point $1$ on the real axis, i.e. the point $P=(1,0).$

Find the nearest point on the circle/exterior to $P=(1,0)$

Since $P$ lies inside the circle or outside? Compute distance from center $C=(2,-1)$ to $P=(1,0)$: $CP=\sqrt{(1-2)^2+(0+1)^2}=\sqrt{1+1}=\sqrt2.$

Since $\sqrt2<\sqrt5,$ the point $P=(1,0)$ lies inside the circle. Therefore, the nearest point in the exterior region $S$ is the point on the boundary along the line from center $C$ to $P$.

Vector from $C$ to $P$ is $(-1,1).$ Its magnitude is $\sqrt2$, so the corresponding unit vector is $\left(-\frac1{\sqrt2},\frac1{\sqrt2}\right).$

Therefore, $z_0 = C + r\cdot \text{(unit vector from $C$ to $P$)}$ $= (2,-1) + \sqrt5\left(-\frac1{\sqrt2},\frac1{\sqrt2}\right).$

Hence $z_0 = 2-\frac{\sqrt{10}}2 + i\left(-1+\frac{\sqrt{10}}2\right).$

Let $z_0=x+iy,$ where $x=2-\frac{\sqrt{10}}2,\qquad y=-1+\frac{\sqrt{10}}2.$

Simplify the required expression

We need the principal argument of $\frac{4-z_0-\overline z_0}{z_0-\overline z_0+2i}.$

Since $z_0+\overline z_0=2x$ and $z_0-\overline z_0=2iy$, we get $4-z_0-\overline z_0 = 4-2x,$ $z_0-\overline z_0+2i = 2iy+2i=2i(y+1).$

Therefore,

Find the principal argument

$-i$ lies on the negative imaginary axis, so its principal argument is $-\frac\pi2.$

Match with options

Therefore the correct option is $\boxed{\text{C }\left(-\frac\pi2\right)}.$

Compare with stored answer

Stored correct answer: C

Our derived answer also gives C. So the stored answer is correct.