Let 𝑆 be the set of all complex numbers z satisfying $$|z^{2} + z + 1| = 1$$. Then which of the following statements is/are TRUE?

Write the quadratic in completed-square form:

$$z^{2}+z+1=\left(z+\frac12\right)^{2}+\frac34$$

Introduce the new variable $$t=z+\frac12$$. The defining condition of the set becomes

$$|t^{2}+\tfrac34|=1 \quad -(1)$$

Equation $$-(1)$$ says that $$t^{2}$$ lies on the circle centred at $$-\,\tfrac34$$ (on the real axis) with radius $$1$$. Hence we may write

$$t^{2}=-\tfrac34+e^{i\phi},\qquad 0\le\phi\lt2\pi$$

The modulus of the right-hand side is

$$m(\phi)=\left|-\tfrac34+e^{i\phi}\right| =\sqrt{\bigl(\cos\phi-\tfrac34\bigr)^{2}+\sin^{2}\phi} =\sqrt{1.5625-1.5\cos\phi}$$

Because $$\cos\phi\in[-1,1]$$ we have

$$0.25\le m(\phi)\le1.75$$

Taking square roots, the modulus of $$t$$ is

$$|t|=\sqrt{m(\phi)}\quad\Longrightarrow\quad 0.5\le|t|\le\sqrt{1.75}\approx1.323\quad -(2)$$

Recall that $$t=z+\tfrac12\;$$, so $$z=t-\tfrac12$$. For any complex number $$t$$ with modulus $$r=|t|$$,

$$|z|=|t-\tfrac12| =\sqrt{\,r^{2}+0.25- r\cos\theta\,},$$

where $$\theta=\arg t$$. For fixed $$r$$ this is maximised when $$\cos\theta=-1$$ (i.e. $$t$$ lies on the negative real axis), giving

$$|z|_{\max}=\sqrt{\,r^{2}+0.25+r\,}$$

Using the largest possible $$r$$ from $$-(2)$$, namely $$r_{\max}=\sqrt{1.75}$$,

$$|z|_{\max}= \sqrt{\,1.75+0.25+\sqrt{1.75}\,} =\sqrt{\,2+\sqrt{1.75}\,}\approx\sqrt{3.322}\approx1.822\lt2$$

Combining these facts we can now test each option.

Option A: $$\bigl|\,z+\tfrac12\,\bigr|\le\tfrac12$$ for all $$z\in S$$.
From $$-(2)$$ we have $$|t|\ge0.5$$, not $$\le0.5$$. Therefore Option A is false. Option B: $$|z|\le2$$ for all $$z\in S$$.
The maximum possible modulus of $$z$$ is about $$1.822$$, which is indeed $$\le2$$. Hence Option B is true. Option C: $$\bigl|\,z+\tfrac12\,\bigr|\ge\tfrac12$$ for all $$z\in S$$.
Inequality $$-(2)$$ gives $$|t|\ge0.5$$ for every admissible $$t$$, so Option C is true. Option D: The set $$S$$ has exactly four elements.
Condition $$|f(z)|=1$$, with $$f$$ a polynomial, defines a one-dimensional real curve in the complex plane; infinitely many $$z$$ satisfy $$|z^{2}+z+1|=1$$. Hence Option D is false.

Therefore the statements that are TRUE are:
Option B and Option C.