<s>[INST] My proof of: $|x - y| < \varepsilon \Leftrightarrow y - \varepsilon < x < y + \varepsilon$ Is it reasonable to prove the following (trivial) theorem?
If yes, is there a better way to do it?

Let $x, y \in \mathbb{R}$.
Let $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$.
$\textbf{Theorem.}$
We have
\begin{equation*}
\left| x - y \right| < \varepsilon
\quad \Longleftrightarrow \quad
y - \varepsilon < x < y + \varepsilon.
\end{equation*}
$\Large \textit{Proof.}$
$\large \textit{Subproof } \Rightarrow .$
$\textbf{Case: } 0 \le x - y . \quad$
Since both
$0 \le x - y$ and
$\left| x - y \right| < \varepsilon$, we have
$x - y < \varepsilon$.
That is, $x < y + \varepsilon$.
The rest of the case proves $y - \varepsilon < x$.
We have $0 \le x - y$.
That is, $y \le x$.
Thus, since $0 < \varepsilon$,
we have $0 + y < \varepsilon + x$.
That is, $y - \varepsilon < x$.
$\textbf{Case: } 0 > x - y . \quad$
Since both
$0 > x - y$ and
$\left| x - y \right| < \varepsilon$, we have
$y - x < \varepsilon$.
That is, $y - \varepsilon < x$.
The rest of the case proves $x < y + \varepsilon$.
We have $0 > x - y$.
That is, $x < y$.
Thus, since $0 < \varepsilon$,
we have $0 + x < \varepsilon + y$.
That is, $x < y + \varepsilon$.
$\large \textit{Subproof } \Leftarrow .$
$\textbf{Case: } 0 \le x - y . \quad$
We have $x < y + \varepsilon$.
That is, $x - y < \varepsilon$.
Also, since $0 \le x - y$, we have $x - y = |x - y|$.
Thus, $|x - y| < \varepsilon$.
$\textbf{Case: } 0 > x - y . \quad$
We have $y - \varepsilon < x$.
That is, $y - x < \varepsilon$.
Also, since $0 > x - y$, we have $y - x = |x - y|$.
Thus, $|x - y| < \varepsilon$.
QED
 [/INST] This is correct. A neater way of putting it would be to note that 
$$
\begin{align}
\mid x-y \mid < \epsilon &\iff x-y \lt \epsilon \text{ and } y-x \lt \epsilon\\
&\iff x-y \lt \epsilon \text{ and } x-y \gt -\epsilon\\
&\iff-\epsilon \lt x-y \lt \epsilon \\
&\iff y-\epsilon \lt x \lt y+\epsilon
\end{align}$$
</s>