Is $C([0,1])$ a "subset" of $L^\infty([0,1])$?

You can actually identify $C([0,1])$ and $C([0,1])/\sim$ because, two continuous fonctions who agree almost everywhere are equal.

Indeed, let $f,g \in C([0,1])$ be such that $A = \{x\in [0,1]\mid f(x) \neq g(x)\}$ is negligible. Then $A$ must have an empty interior, so its complementary is dense in $[0,1]$. The function $h = f-g$ is continuous, hence $h([0,1]) = h(\overline{[0,1]\setminus A}) \subset \overline{h([0,1]\setminus A)} = \{0\}$. This proves that $f=g$.

If you want to be really rigorous, it would be better to say that the natural injection $C([0,1]) \hookrightarrow \mathcal{L}^\infty([0,1])$ factorizes with $\sim$, so that it induces an injection $C([0,1]) \hookrightarrow L^\infty([0,1])$. That way, $C([0,1])$ is identified with the image of this injection.