The largest value of $$r$$, for which the region represented by the set $$\{\omega \in C | |\omega - 4 - i| \leq r\}$$ is contained in the region represented by the set $$\{z \in C | |z - 1| \leq |z + i|\}$$, is equal to:

Let us denote the required closed disc by

$$\{\;\omega\in\mathbb C\;|\;|\omega-(4+i)|\le r\;\}.$$

Its centre is the complex number $$4+i,$$ that is, the point $$\bigl(4,\,1\bigr)$$ in the Argand plane, and its radius is the real number $$r$$ that we have to determine.

The containing region is

$$\{\,z\in\mathbb C\;|\;|z-1|\le|\,z+i\,|\ \},$$

i.e. the set of all points whose distance from $$1$$ is not larger than the distance from $$-i$$. We rewrite the inequality in Cartesian form. Take a general point $$z=x+iy$$; then

$$|z-1|^2=(x-1)^2+y^2,$$ $$|\,z+i\,|^2=x^2+(y+1)^2.$$

Imposing equality gives the boundary of the region:

$$|z-1|^2=|\,z+i\,|^2 \;\Longrightarrow\; (x-1)^2+y^2=x^2+(y+1)^2.$$

Expanding every square, we obtain

$$x^2-2x+1+y^2=x^2+y^2+2y+1.$$

After cancelling the common terms $$x^2,\;y^2,\;1$$ on both sides, we arrive at

$$_{-2x}=_{2y}\quad\Longrightarrow\quad y=-x.$$

Thus the boundary is the straight line $$y=-x,$$ or equivalently $$x+y=0.$$ Because the inequality is “$$\le$$”, the desired region includes this line.

Now we must decide on which side of the line the region lies. We test the point $$z=1$$ (the complex number $$1+0i$$) because it obviously satisfies the original inequality: $$|1-1|=0,\qquad |1+i|=\sqrt2,\qquad 0\le\sqrt2.$$ For $$z=1$$ we have $$x=1,\;y=0,$$ so $$x+y=1>0.$$ Hence the region is the half-plane

$$x+y\ge0.$$

The geometric problem has now become: inscribe the largest possible closed circle with centre $$C(4,1)$$ inside the closed half-plane $$x+y\ge0.$$ The radius of this largest circle is simply the perpendicular distance from the centre to the bounding line $$x+y=0.$$

We recall the distance formula from a point $$(x_0,y_0)$$ to the line $$ax+by+c=0$$:

$$\text{Distance}=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.$$

For the line $$x+y=0$$ we have $$a=1,\;b=1,\;c=0.$$ For the point $$C(4,1)$$ we substitute $$x_0=4,\;y_0=1.$$

We obtain

$$\text{Distance} =\dfrac{|1\cdot4+1\cdot1+0|}{\sqrt{1^2+1^2}} =\dfrac{|5|}{\sqrt2} =\dfrac{5}{\sqrt2} =\dfrac{5\sqrt2}{2}.$$

This distance is the largest possible radius $$r$$ such that the whole disc remains inside the half-plane. Therefore

$$r=\dfrac{5\sqrt2}{2}.$$

Looking at the given options, we see that this value corresponds to Option D.

Hence, the correct answer is Option D.