Let $S=\left\{z=x+iy:\left|z-1+i\right|\ge \left|z\right|,\left|z\right|<2,\left|z+i\right|=\left|z-1\right|\right\}$.  Then the set of all values of $x$, for which $w=2x+iy\in S$ for some $y\in \mathrm{ℝ}$, is

Given,

$\left|z-1+i\right|\ge \left|z\right|, \left|z\right|<2 \& \left|z+i\right|=\left|z-1\right|$

Now solving $\left|z-1+i\right|\ge \left|z\right|$ put $z=x+iy$ we get, $x-y\le 1$ so plotting we get,

Now solving $\left|z\right|<2$ we get, $x^2+y^2<4$ on plotting the diagram we get,

Now solving $\left|z+i\right|=\left|z-1\right|$ we get $x+y=0$

Now combine diagram of $x+y=0 \& x-y\le 1$ will be,

Now combining all diagram we get,

Hence

$w=2x+iy\in S$

Now for $w\in S$, $2x\le \frac{1}{2}$ so $x\le \frac{1}{4}$

Now $2x+y=0\Rightarrow y=-2x$ putting the value in $x^2+y^2<4$

$\Rightarrow {\left(2x\right)}^2+{\left(2x\right)}^2<4$

$\Rightarrow x^2<\frac{1}{2}\Rightarrow x\in \left(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

So, $x\in \left(\frac{-1}{\sqrt{2}},\frac{1}{4}\right]$