Let $$h\left(x\right) = \text{min}\left\{|\sin x|, | \cos x|\right\}$$, for all real numbers $$x$$. Let S be the set of points in $$\left(0, \frac{\pi}{2}\right)$$ where $$h(x)$$ is not differentiable. Then the cardinality of S is:

$$h\left(x\right) = \text{min}\left\{|\sin x|, | \cos x|\right\}$$
$$x\in\ $$ $$\left(0, \frac{\pi}{2}\right)$$
Since, both $$sinx$$ and $$cosx$$ are positive in the interval $$\left(0, \frac{\pi}{2}\right)$$ 
$$h\left(x\right) = \text{min}\left\{\sin x,  \cos x\right\}$$
We know that 
$$\sin x<\cos x$$in the interval $$\left(0, \frac{\pi}{4}\right)$$
$$x=\frac{\pi\ }{4}\ $$, $$sinx = cosx$$
$$\sin x>\cos x$$ in the interval $$\left(\ \frac{\ \pi\ }{4},\ \frac{\ \pi\ }{2}\right)$$
$$h\left(x\right)$$ will take the value $$sinx$$ in the interval $$\left(0, \frac{\pi}{4}\right)$$ and the value $$cosx$$ in the interval $$\left(\ \frac{\ \pi\ }{4},\ \frac{\ \pi\ }{2}\right)$$
Differentiating wrt $$x$$,
$$h'\left(x\right)=\cos\ x,\ \ x\in\ \left(0,\ \frac{\ \pi\ }{4}\right)$$
$$h'\left(x\right)=-\sin\ x,\ \ x\in\ \left(\ \frac{\ \pi\ }{4},\ \frac{\ \pi\ }{2}\right)$$
At $$x=\frac{\ \pi\ }{4}$$ the values of $$cosx$$ and $$-sinx$$ will not be equal.
The function $$h\left(x\right)$$ is continuous and differentiable elsewhere.
Hence, the cardinality of S is 1.