Let $w \in \mathbb{C}:|w| \leq r\land x > \sqrt{2}r$ for a given $r > 0$. I am trying to understand how that gives the inequality $\left|\frac{2w}{w^2 - x^2}\right|\leq\frac{4r}{x^2}$ but my geometric/algebraic perception is not good enough.
One can immediately write out that
$$\left|\frac{2w}{w^2 - x^2}\right|\leq \frac{2r}{|(w - x)(w + x)|}$$
and I have a feeling that you would need some geometric fact to conclude the given claim. Conceptually what we have is the product of the distances from the symmetric points $(0, x), (0, -x)$ to the points of the closed disk $\overline{B_r(0)}$ with $x > \sqrt{2}r$.
Just use the triangle inequality to estimate the denominator: $$ |w^2-x^2| \ge x^2 - |w|^2 \ge x^2 - r^2 \ge x^2 - \frac 12 x^2 = \frac 12 x^2 > 0\, . $$ It follows that $$ \left|\frac{2w}{w^2 - x^2}\right|\leq \frac{2r}{\frac 12 x^2} = \frac{4r}{x^2} \, . $$