\usepackage{mathtools}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}

The command \ceil will do; if called as \ceil*{x} it will add \left and \right; you can also call it as

\ceil[\big]{x} \ceil[\Big]{x} \ceil[\bigg]{x} \ceil[\Bigg]{x}

to state explicitly the size of the delimiters.

I didn't read through your proof, but I would do it this way: can be expressed as

where , with and . There are 3 cases:

1. and : then both sides are clearly equal to .

2. and :

3. :

Your function scales both input $x$ and output $y$ up by a factor of $100$: $$ \lfloor 100 x \rfloor = 100y, $$ i.e. if we define these scaled coordinates $X = 100x$ and $Y = 100y$, then the equation relating inputs to outputs looks like $$ \lfloor X \rfloor = Y, $$ which you can think of the prototype of the relationship between the variables. Putting all the transformations in one diagram looks like
$$ x \to X \to Y \leftarrow y, $$ so the only way to build the composition $x \to y$ is to invert that last arrow $Y \leftarrow y$ to produce $Y \to y$, namely to divide by the scaling factor, hence we introduce fractions.

As always, we can hide the fraction inside of a definition, which doesn't change the fact that we're dividing by the scaling factor, but cosmetically, it might look nicer. We could, for instance, for any $p > 0$, define a rounding function with precision $\frac1p$ by $$ \operatorname{floor}_p(x) = \frac1p \lfloor p x \rfloor $$ that has the property that $\operatorname{floor}_p(x) = y$ for all $y \leq x \leq y + \frac1p$. With this notation, your example would look like $$ \operatorname{floor}_{100}(2.4783) = 2.47. $$