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

We have $$S = \{z = x + iy : |z-1+i| \geq |z|,\; |z| < 2,\; |z+i| = |z-1|\}$$, and we need $$w = 2x + iy \in S$$ for some $$y \in \mathbb{R}$$. We must find the set of all such values of $$x$$.

Since $$w = 2x + iy \in S$$, we write $$w = u + iv$$ where $$u = 2x$$ and $$v = y$$. The conditions on $$w$$ are:

Condition 1: $$|w + i| = |w - 1|$$, i.e., $$|u + i(v+1)| = |(u-1) + iv|$$. Squaring: $$u^2 + (v+1)^2 = (u-1)^2 + v^2$$, which simplifies to $$u^2 + v^2 + 2v + 1 = u^2 - 2u + 1 + v^2$$, giving $$2v = -2u$$, so $$v = -u$$. Thus $$y = -2x$$.

Condition 2: $$|w| < 2$$, i.e., $$u^2 + v^2 < 4$$. Since $$v = -u$$, we get $$u^2 + u^2 < 4$$, so $$2u^2 < 4$$, meaning $$u^2 < 2$$, i.e., $$-\sqrt{2} < u < \sqrt{2}$$. Since $$u = 2x$$, we have $$-\dfrac{1}{\sqrt{2}} < x < \dfrac{1}{\sqrt{2}}$$.

Condition 3: $$|w - 1 + i| \geq |w|$$, i.e., $$|(u-1) + i(v+1)| \geq |u + iv|$$. Squaring: $$(u-1)^2 + (v+1)^2 \geq u^2 + v^2$$, which gives $$-2u + 1 + 2v + 1 \geq 0$$, so $$-2u + 2v + 2 \geq 0$$, i.e., $$v \geq u - 1$$. Substituting $$v = -u$$: $$-u \geq u - 1$$, so $$1 \geq 2u$$, giving $$u \leq \dfrac{1}{2}$$, i.e., $$x \leq \dfrac{1}{4}$$.

Combining: $$-\dfrac{1}{\sqrt{2}} < x \leq \dfrac{1}{4}$$.

Hence, the correct answer is Option B: $$\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{4}\right)$$.