Is $C([0,1])$ a "subset" of $L^\infty([0,1])$? This is motivated from an exercise in real analysis:

Prove that $C([0,1])$ is not dense in $L^\infty([0,1])$.

My first question is how $C([0,1])$ is identified as a subset of $L^\infty([0,1])$? (I think one would never say something like "$A$ is (not) dense in $B$" if $A$ is not even a subset of $B$. )
First of all, $L^\infty([0,1])$ is defined as a quotient space, but $C([0,1])$ is a set of functions:
$$
C([0,1]):=\{f:[0,1]\to{\Bbb R}\mid f \  \text{is continuous}\}. \tag{1}
$$
I think one should also take $C([0,1])$ as 
$$
C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \  \text{for some g where g is continuous on}\ [0,1]\} \tag{2}
$$
where $f\sim g$ if only if $f=g$ almost everywhere. But I've never read any textbook (PDE, measure theory, or functional analysis, etc) that defines $C([0,1])$ (or more generally $C(X)$ where $X\subset{\Bbb R}$ is compact) in this way before. 
Second question: Could anyone come up with a reference with such definition?

[EDITED:]The original title doesn't reflect my point. I've changed it accordingly. 
[EDITED:] Some thoughts after reading the comments and answers:
When one regards $C([0,1])$ as  a subset of $L^\infty([0,1])$, (1) is  not correct, and (2) would be not correct either. The final version I can come up with is 
$$
C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \  \text{for some g where g is continuous on}\ [0,1]\}\big/\sim.
\tag{3}
$$  
 A: 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.
A: This includes an answer to the original question posted, modified to answer the latest version of the question at time of writing. I have left the original answer because it includes an observation (and an example of how that observation is applied) that  illustrates why people are a bit careless about differentiating between the equivalence classes and their representatives. 
The norm on $L^\infty[0,1]$ is the essential supremum, so it 'ignores' changes on null sets. By $\|f\|$ below, I mean the essential supremum norm.
I use $[f]$ to mean the equivalence class of $f$, the notation is potentially confusing, but context will disambiguate. By $f_1 \sim f_2$, I mean that $\{x | f_1(x) \neq f_2(x) \}$ is a null set.
One identifies $C[0,1]$ with a subset of $L^\infty[0,1]$ by taking equivalence classes, ie, we are really dealing with $\{[f] | f \in C[0,1] \}$, which is a subset of $L^\infty[0,1]$. (As an aside, continuity means that the identification $f \mapsto [f]$ is injective.)
(I use $m$ below to denote the Lebesgue measure, however the observation holds for any measure, of course. The subsequent demonstration of 'not being dense' does depend on the Lebesgue measure.)
Observation: Suppose $P$ is some property on $\mathbb{R}$ (or $\mathbb{C}$ as the case may be), and suppose $f_1 \sim f_2$. Then we have $m \{x | P(f_1(x)) \} = m \{x | P(f_2(x)) \}$. This follows since $f_1(x) = f_2(x)$  a.e. $x$. So even though $[f]$ is an equivalence class, we can think about $m \{x | P(\,[f]\,(x)) \}$ with the understanding that we really mean $m \{x | P(f(x)) \}$ for some representative $f \in [f]$. This is what allows us to be somewhat blasé about dealing with a function vs. its equivalence class.
The previous observation can be extended considerably, but loosely the idea is that the measure of the set of points that satisfies a 'nice' property is independent of the particular representations from the equivalence classes. By a 'nice' property, I mean a property whose truth value at $x$ depends only on the values of the representations at $x$.
Now consider $ [1_{[\frac{1}{2},1]}]$, and $[c]$ where $c \in C[0,1]$.
I claim $\|[1_{[\frac{1}{2},1]}]-[c] \| \ge \frac{1}{2}$, and since $c$ was arbitrary, we see that $C[0,1]/ \sim$ cannot be dense in $L^\infty[0,1]$.
To see why the claim is true, we will prove the statement for specific representatives of $[1_{[\frac{1}{2},1]}], [c]$ (ie, $1_{[\frac{1}{2},1]}, c$ respectively) and then invoke the above observation to conclude. 
Let $\gamma =c(\frac{1}{2})$. We have $|\gamma-1|+|\gamma| \ge 1$, and hence 
$\max(|\gamma-1|, |\gamma|) \ge \frac{1}{2}$. Continuity of $c$ implies that for any $\epsilon>0$, there is a $\delta >0$ such that $|c(x)-\gamma| < \epsilon$ whenever $x \in B(\frac{1}{2}, \delta)$. Noting that if $x \in B(\frac{1}{2}, \delta)$, we have $1-x \in B(\frac{1}{2}, \delta)$, we get $\max(|c(1-x)|, |c(x)-1|) \ge \max(|\gamma|-\epsilon, |\gamma-1|-\epsilon) \ge \frac{1}{2}-\epsilon$. Hence for $x\in (\frac{1}{2},\frac{1}{2}+\delta)$, we have $\max(|c(1-x)-1_{[\frac{1}{2},1]}(1-x)|, |c(x)-1_{[\frac{1}{2},1]}(x)|) \ge  \frac{1}{2}-\epsilon$.
In particular, $m \{x |\  |c(x)-1_{[\frac{1}{2},1]}(x)| \ge  \frac{1}{2}-\epsilon \} \ge \delta >0$. The above observation shows that this is true for all $c'\in [c], f'\in [1_{[\frac{1}{2},1]}]$, so it follows from the definition of essential supremum that $\|[c]-[1_{[\frac{1}{2},1]}]\| \ge \frac{1}{2}$.
A: Actually, such identifications are done in the textbook about partial differential equations and Sobolev spaces. For example, we can see theorems like "$C_0^\infty(\Bbb R^d)$, the set of smooth functions with compact support, is dense in $W^{1,p}(\Bbb R^d)$ for all $1\leqslant p<\infty$". This means that we identify a test function $\phi$ to the class of functions which are almost everywhere equal to $\phi$, as what is done in the OP.
The characteristic function $g$ of $[1/2,1)$ cannot be approached in $L^\infty$ by such functions, because there would be an $f$ almost everywhere equal to a continuous function such that $\sup_{x\in [0,1]\setminus N}|g(x)-f(x)|<1/3$, where $N$ is a set of $0$ measure. In particular, $\sup_{x\in [0,1/2)\setminus N}|f(x)|<1/3$ and $\sup_{x\in [1/2,1)\setminus N}|1-f(x)|<1/3$.
