The maximum value of ${\cos }^2\left(\frac{\pi }{4}-x\right)+{\left(\sin x-\cos x\right)}^2$  is

$=\frac{1}{2}\left(1+\sin 2x\right)+1-\sin 2x$

$=\frac{3}{2}-\frac{1}{2}\sin 2x$

$-1\le \sin 2x\le 1$

$-\frac{1}{2}\le \frac{1}{2}\sin 2x\le \frac{1}{2}$

$\frac{1}{2}\ge -\frac{1}{2}\sin 2x\ge -\frac{1}{2}$

$2\ge \frac{3}{2}-\frac{1}{2}\sin 2x\ge 1\left(\because \text{Add }\frac{3}{2}\text{on both sides}\right)$