Suppose that $z\in\mathbb C$ with $|z^2+1|\le 1$. How to prove $|z+1|\geq\frac12$.

Let $z=a+ib$, then $$|1+z^2|^2=1+2a^2+a^4+b^2(2a^2-2)+b^4\le1.$$ We want to minimize $|1+z|^2=\color{orange}{(1+a)^2+b^2}$. Let me complete the square in the constraint: $$b^4+b^2(2a^2-2)\color{blue}{+(a^2-1)^2}\le\color{blue}{(a^2-1)^2}-a^4-2a^2=1-4a^2$$ which is the same as $$(b^2+(a^2-1))^2\le1-4a^2.$$ In particular, it follows that $- \frac12<a<\frac12$ and $$b^2\in\left[1-a^2-\sqrt{1-4a^2},1-a^2+\sqrt{1-4a^2}\right].$$

Thus, $$2+2a-\sqrt{1-4a^2}\le\color{orange}{(1+a)^2+b^2}\le2+2a+\sqrt{1-4a^2}.$$ We will thus want to minimize the function $2a-\sqrt{1-4a^2}$. This can be done for example with calculus, but we can also proceed as follows:

Now we have $$8a^2-4 \sqrt 2 a+1=8 \left(a - \frac{1}{2 \sqrt 2}\right)^2\geq0$$ which implies $$4a^2-4\sqrt 2a+2=(\sqrt 2-2a)^2\geq1-4a^2$$ and thus $$\sqrt 2-2a\geq\sqrt{1-4a^2}.$$ By replacing $a$ by $-a$ we also get $$\sqrt{1-4a^2}\le\sqrt2+2a.$$

It follows from the previous result that $$2-\sqrt 2\le\color{orange}{(1+a)^2+b^2}\le2+\sqrt 2$$ and thus in particular $$\bbox[15px,border:1px groove navy]{|1+z|=\sqrt{\color{orange}{(1+a)^2+b^2}}\geq\sqrt{2-\sqrt 2}>\frac12.}$$

The curves that satisfy

$$|z^2+1| = 1,\>\>\>\>\>|z+1| = \frac12\tag 1$$

can be expressed in polar coordinates $z=re^{i\theta}$ as

$$r^4+2r^2\cos2\theta=0, \>\>\>\>\> r^2+2r\cos\theta + \frac34=0 $$

Eliminate the variable $\theta$ to obtained the following quadratic equation in $r^2$,

$$2r^4-\frac12r^2+\frac9{16}=0$$

Its discriminate is negative, meaning the two curves given by (1) have no intersections. (See the plot below.)

It is straightforward to verify that $z=-1$, the center of the circle $|z+1| = \frac12$, does not satisfies the $|z^2+1| \le 1$, which means the circle's center lies outside the region $|z^2+1| \le 1$. Then, along with the earlier conclusion that the circle does not intersect the other curve, we further conclude that the circle itself $|z+1| = \frac12$ is also outside the given region $|z^2+1| \le 1$. Thus,

$$|z+1|>\frac12$$