We know that the complex number $z$ satisfying $\left|z-z_0\right|=r$ represents a circle with centre $z_0$ and radius $r$ units.

Hence, for $S_1=|z-1|\le \sqrt{2}, z$ lies on and inside the circle
of radius $\sqrt{2}$ units and centre $(1, 0).$

For $S_2,$ let $z=x+iy$

Now, $(1-i)\left(z\right)=(1-i)(x+i y)$

$\Rightarrow (1-i)\left(z\right)=x+iy-ix-i^{2}y$

$\Rightarrow (1-i)\left(z\right)=x+iy-ix+y$

$\Rightarrow Re\left(\right(1-i\left)z\right)=x+y$

$\Rightarrow x+y\ge 1.$

And, for $S_3,$ again let $z=x+iy,$

$\Rightarrow y\le 1.$

Plotting all the inequalities on the graph, we get

Now, the common part is shown in the shaded part, hence, we get infinite points in the shaded region.

$\Rightarrow S_1\cap S_2\cap S_3$ has infinitely many elements.