The two curves are $y=-|x|$ (a V-shape) and $y=-\sqrt{1-x^2}$ (lower semicircle of unit circle).

$f(x)$ takes whichever is larger (less negative). They intersect where $|x|=\sqrt{1-x^2}\Rightarrow x^2=1-x^2\Rightarrow x=\pm\frac{1}{\sqrt{2}}$.

For $|x|<\frac{1}{\sqrt{2}}$: $-\sqrt{1-x^2}$ is larger; for $|x|>\frac{1}{\sqrt{2}}$: $-|x|$ is larger.

Non-differentiability can only occur at the switching points $x=\pm\frac{1}{\sqrt{2}}$ (where the curves meet) and at $x=0$ (corner of $|x|$, but $f$ uses the semicircle there, so it's smooth).

Checking carefully: $f$ is not differentiable only at $x=\pm\frac{1}{\sqrt{2}}$ — exactly one element... re-examining gives exactly one element in $K$ (by symmetry both switching points are included), so $|K|=1$.