Given $x,y \in \mathbb{R}$, we define $x \sim y$ if $x - y \in \mathbb{Z}$, so $\mathbb{R}/\mathbb{Z}$ is the set of equivalence classes, and I want to prove that the set of equivalence classes is the set $[0,1)$. Here is my attempt.
Given $x \in \mathbb{R}$, define the floor of $x$, $\lfloor x \rfloor$ to be the smallest integer less than or equal to $x$. By definition, we have $\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1$, so $0 \leq x - \lfloor x \rfloor < 1$, so there exists $y \in [0,1)$ such that $x \sim y$. That is, every $x \in \mathbb{R}$ lives in the equivalence class of some element of $[0,1)$, where we simply write $x = s + t$ where $s = \lfloor x \rfloor$ is the integer component and $t = x - \lfloor x \rfloor$ is the fractional component, and we have $x \sim t$. Furthermore, no two elements of $[0,1)$ belong to the same equivalence class. Indeed, if $a,b$ are distinct elements of $[0,1)$, then $a-b \in (-1,1) \setminus \{0\}$, so $a-b \not \in \mathbb{Z}$ and hence $a \not \sim b$. Therefore, the set $\mathbb{R}/\mathbb{Z}$ of equivalence classes is exactly the interval $[0,1)$.
The thing I'm mostly unsure of the need to prove this fact in both directions. I need to show that every element of $[0,1)$ is an equivalence class, that these are all the equivalence classes, and that this description of equivalence classes doesn't include any overlap. I proved that every element belongs to some equivalence class $[0,1)$ and that there isn't any overlap, but I don't think I proved that each element of $[0,1)$ is an equivalence class. I don't know exactly how to prove that, other than to say that if I fix $r \in [0,1)$ and find $x \in \mathbb{R}$ such that $x - \lfloor x \rfloor = r$.
I'd be very interested on some feedback on the proof and on the appropriate sequencing of steps.