This morning a colleague asked me how I would solve quickly this limit, lim x...

Consider rewriting the numerator as $\cos (2x)-\cos (\frac{\pi }{2}).$ Then we can use the same manipulation to get
$\cos (2x)-\cos (\frac{\pi }{2})=-2\sin \left(\frac{2x+\frac{\pi }{2}}{2}\right)\sin \left(\frac{2x-\frac{\pi }{2}}{2}\right)=-2\sin (x+\frac{\pi }{4})\sin (x-\frac{\pi }{4})$
Substituting this in gives us
$\underset{x\to \frac{\pi }{4}}{lim}\frac{-2\sin (x+\frac{\pi }{4})\sin (x-\frac{\pi }{4})}{2(-2\sin \left(\frac{x+\frac{\pi }{4}}{2}\right)\sin \left(\frac{x-\frac{\pi }{4}}{2}\right))}=\frac{1}{2}\underset{x\to \frac{\pi }{4}}{lim}\frac{\sin (x+\frac{\pi }{4})}{\sin \left(\frac{x+\frac{\pi }{4}}{2}\right)}\cdot \underset{x\to \frac{\pi }{4}}{lim}\frac{\sin (x-\frac{\pi }{4})}{\sin \left(\frac{x-\frac{\pi }{4}}{2}\right)}$
supposing for the moment that those two limits exist. (also note the additional factor of 2 in the denominator which seems to have been dropped in your manipulation)
For the first limit we can simply evaluate at $x=\frac{\pi }{4}$ to get $\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}.$
For the second limit, consider rewriting $\sin (x-\frac{\pi }{4})$ in the numerator as $\sin \left(2\frac{x-\frac{\pi }{4}}{2}\right)=2\sin \left(\frac{x-\frac{\pi }{4}}{2}\right)\cos \left(\frac{x-\frac{\pi }{4}}{2}\right).$ This gives us that
$\underset{x\to \frac{\pi }{4}}{lim}\frac{\sin (x-\frac{\pi }{4})}{\sin \left(\frac{x-\frac{\pi }{4}}{2}\right)}=\underset{x\to \frac{\pi }{4}}{lim}2\cos \left(\frac{x-\frac{\pi }{4}}{2}\right)=2$
So our total limit is $\frac{1}{2}\cdot \sqrt{2}\cdot 2=\sqrt{2},$ which agrees with your initial solution.

If you set $x=u+\frac{\pi }{4}$ with $u\to 0$ then you get
$f(x)={\displaystyle \frac{\cos (2x)}{2\cos (x)-\sqrt{2}}}={\displaystyle \frac{-2\sin (u)\stackrel{\to 1}{\overbrace{\cos (u)}}}{\sqrt{2}(\underset{\ll \sin (u)}{\underbrace{\cos (u)-1}}-\sin (u))}}\sim {\displaystyle \frac{-2\sin (u)}{-\sqrt{2}\sin (u)}}\to \sqrt{2}$
The fact that $\cos (u)-1\ll \sin (u)$ can be inferred from:
$\left|{\displaystyle \frac{\cos (u)-1}{\sin (u)}}\right|=\left|{\displaystyle \frac{2\sin (u/2{)}^2}{\sin (u)}}\right|=\underset{\to 1}{\underbrace{\left|{\displaystyle \frac{\sin (u/2{)}^2}{u^2/4}}\right|}}\times \underset{\to 1}{\underbrace{\left|{\displaystyle \frac{u}{\sin (u)}}\right|}}\times \underset{\to 0}{\underbrace{|8u|}}\to 0$