$| \dfrac{-z+\sqrt{z^2-4}}{2}|\le 1$ Suppose $z=u+iv$, with $v>0$.
$$ m(z)=\dfrac{-z+\sqrt{z^2-4}}{2}$$ satisfies the equation $$m(z)+1/m(z)+z=0 $$ from which we found that $$|m(z)|=|m(z)+z|^{-1} $$
Take the branch of the square root so that the imaginary part of $m(z)$ is postive.
($m(z) $ here is the Stieljes Transformation of the Semicircle Law)
How do we show $$|m(z)|\le 1 $$ or equivalently $$|m(z)+z|\ge 1 $$

As pointed out by Conrad in the comment,$$|m(z)(z+m(z))|=1 $$If we can show $m(z)$ is the smaller of the two in norm, then we are done.
Noting that $$v>0,\qquad m(z)=\dfrac{-z+\sqrt{z^2-4}}{2},\qquad m(z)+z= \dfrac{z+\sqrt{z^2-4}}{2}$$
we have $$Im[m(z)+z] >Im[m(z)] $$ but how about the real part?

as shown in the first few lines of the accepted solution, by the constrains we have on the imaginary part of $z$ and $m(z)$. we have that the real and imaginary part of $$z$$ and $$\sqrt{z^2-4}$$ are of the of the sign. With this we can conclude that $$|m(z)+z|\ge |m(z)| $$

Let $z=a+ib,\sqrt{z^2-r^2}=c+id$ where $a,b,c,d$ are real and $r>0$
$$r^2=z^2-(z^2-r^2)=(a+ib)^2-(c+id)^2=a^2+d^2-b^2-c^2+2i(ab-cd)$$
Equating the imaginary parts, $ab-cd=0\implies\dfrac ac=\dfrac db=k$(say)
$\implies a=ck,d=bk$
Equating the real parts,  $$r^2=a^2+d^2-b^2-c^2=(b^2+c^2)(k^2-1)$$
$\implies k^2-1=\dfrac{r^2}{b^2+c^2}>0$
$\implies$ either $k>1$ or $k<-1$
$$|-z+\sqrt{z^2-r^2}|^2=|c+id-(a+ib)|^2=(c-a)^2+(d-b)^2=(b^2+c^2)(1-k)^2$$
$$=\dfrac{r^2(1-k)^2}{(k^2-1)}=\dfrac{r^2(k-1)}{k+1}$$ which will be $\le r^2$
$\iff\dfrac{k-1}{k+1}\le1\iff0\ge\dfrac{k-1}{k+1}-1=-\dfrac2{k+1}$
$\iff 0\le\dfrac2{k+1}\iff k+1>0\iff k>-1$
So, the proposition will be nullified if $k<-1$
