$f\in C^1([a,b];\Bbb R)$ piecewise $\iff\ \exists\sigma= \{a=a_0<a_1<\dots<a_n=b\}$ a partition of $[a,b]\text{ such that } f'|_{(a_i,a_{i+1})}\in C^0$ $\forall i\in\{0,\dots,n-1\}$ and $f\in C^0$
The function $x\mapsto\sqrt{|x|}$ is not $C^1([-1,1];\Bbb R)$ piecewise because the derivative approaches infinity at zero. But $\frac{1}{2\sqrt{x}}|_{(0,1)}$ is continuous. The only problem is that it is not extendable by continuity to $[0,1)$.
Is that part of the definition of $f'|_{(a_i,a_{i+1})}\in C^0$ i.e. does $f'|_{(a_i,a_{i+1})}\in C^0$ imply that $f'|_{(a_i,a_{i+1})}$ should have a finite limit in $a_i$ and $a_{i+1}$?