In Brezis' book (chapter 8, section 2), he mentioned that a continuous function on $[0,1]$ that is piecewise $C^1$ on $[0,1]$ belongs to $H^1(0,1)$.
Shouldn't we restrict to $C(0,1)$? or because every element in $H^1$ is an equivalent class of $L^2(0,1)$ and we perform restriction on the $C[0,1]$ to $C(0,1)$, so the end point $0,1$ does not matter here?
Note that $C^1(0,1)$ is a larger space than $C^1[0,1]$, since the map $f\mapsto f|_{(0,1)}$ is injective, so if you could say that $C^1(0,1)$ is a subspace of $H^1(0,1)$, you'd be in the clear. Note, however, that not all $C^1$ functions on $(0,1)$ admit $L^2$derivatives. Indeed, take any function $g\in (C(0,1)\cap L^1(0,1))\setminus L^2(0,1)$ and consider $f(x)=\int_0^x g(t)\textrm{d}t$.
So what we really need is some assurance that the classical derivatives will also be integrable. This is true for $C^1[0,1]$ functions since it is true for the class of $C^1$ functions on $(0,1)$ with classical derivatives in $L^{\infty}$. Hence, the latter is really the "natural class at play here".