My proof of: $|x - y| < \varepsilon \Leftrightarrow y - \varepsilon < x < y + \varepsilon$

Question

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

Answer 1

This is correct. A neater way of putting it would be to note that $$ \begin{align} \mid x-y \mid < \epsilon &\iff x-y \lt \epsilon \text{ and } y-x \lt \epsilon\\ &\iff x-y \lt \epsilon \text{ and } x-y \gt -\epsilon\\ &\iff-\epsilon \lt x-y \lt \epsilon \\ &\iff y-\epsilon \lt x \lt y+\epsilon \end{align}$$