Let $z=x+iy$, where $x,y\in\mathbb{R}$.

We convert each set into its geometric form in the Argand plane.

1. Interpret $S_1$

Given $S_1=\{z\in\mathbb{C}:|z-1|\le \sqrt2\}.$

Now, $|z-1|=|x+iy-1|=|(x-1)+iy|,$ so $(x-1)^2+y^2\le 2.$

Thus $S_1$ is the closed disk centered at $(1,0)$ with radius $\sqrt2$.

2. Interpret $S_2$

Given $S_2=\{z\in\mathbb{C}:\operatorname{Re}((1-i)z)\ge 1\}.$

Compute: $(1-i)z=(1-i)(x+iy)=x+iy-ix-i^2y=(x+y)+i(y-x).$

Hence, $\operatorname{Re}((1-i)z)=x+y.$

So $x+y\ge 1.$

Thus $S_2$ is the half-plane on or above the line $x+y=1.$

3. Interpret $S_3$

Given $S_3=\{z\in\mathbb{C}:\operatorname{Im}(z)\le 1\}.$

Since $\operatorname{Im}(z)=y$, this is simply $y\le 1.$

So $S_3$ is the half-plane on or below the line $y=1$.

4. Find the intersection geometrically

We need points satisfying all three: $(x-1)^2+y^2\le 2, \qquad x+y\ge 1, \qquad y\le 1.$

This is the intersection of a closed disk with two closed half-planes. Such an intersection, if nonempty and containing a region/arc/segment, will have infinitely many points.

So let us check whether it is nonempty and not just a finite set.

Take the point $z=1 \quad (x=1,y=0).$

Check:

1. For $S_1$: $|1-1|=0\le \sqrt2.$
2. For $S_2$: $x+y=1+0=1\ge 1.$
3. For $S_3$: $y=0\le 1.$

So $z=1$ belongs to the intersection.

Now check nearby points on the line $y=0$ with $x\ge 1$.

Let $z=x$ with $y=0$. Then conditions become:

• From $S_1$: $(x-1)^2\le 2 \quad \Rightarrow \quad 1-\sqrt2\le x\le 1+\sqrt2.$
• From $S_2$: $x+0\ge 1 \Rightarrow x\ge 1.$
• From $S_3$: $0\le 1,$ which is always true.

Hence every real number $x$ in the interval $1\le x\le 1+\sqrt2$ corresponds to a point in $S_1\cap S_2\cap S_3$.

This is an entire line segment, so the intersection contains infinitely many elements.

5. Evaluate options

• A: has exactly three elements — false
• B: is a singleton — false
• C: has infinitely many elements — true
• D: has exactly two elements — false

Therefore, the correct option is $\boxed{\text{C}}.$