Is my proof correct?
Let $(x_n)_{ n \in \mathbb{N} }$ be a real sequence.
$\textbf{Definition 1.}$ $(x_n)$ is $\textit{convergent}$ iff there is an $x \in \mathbb{R}$ such that, for every $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$, there is an $N \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$ with $n > N$, we have $|x_n - x| < \varepsilon$.
$\textbf{Definition 2.}$ $(x_n)$ is a $\textit{Cauchy sequence}$ iff, for every $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$, there is an $N \in \mathbb{N}$ such that, for every $m,n \in \mathbb{N}$ with $m,n > N$, we have $|x_m - x_n| < \varepsilon$.
$\textbf{Theorem.}$ If $(x_n)$ is convergent, then it is a Cauchy sequence.
$\textit{Proof.}$ We assume that $(x_n)$ is convergent. It remains to prove that $(x_n)$ is a Cauchy sequence. To that end, we consider Definition 2. Let $\varepsilon \in \mathbb{R}$ such that $\varepsilon > 0$. It remains to prove that there is an $N \in \mathbb{N}$ such that, for every $m,n \in \mathbb{N}$ with $m,n > N$, we have $|x_m - x_n| < \varepsilon$. Therefore, our next step is to choose $N$. Since $(x_n)$ is convergent, by Definition 1, there are $x \in \mathbb{R}, N' \in \mathbb{N}$ such that, for every $n' \in \mathbb{N}$ with $n' > N'$, we have \begin{equation*} |x_{n'} - x| < \varepsilon / 2. \qquad\qquad (\textbf{Property 1}) \end{equation*} We choose $N := N'$. It remains to prove that, for every $m,n \in \mathbb{N}$ with $m,n > N$, we have $|x_m - x_n| < \varepsilon$. Let $m,n \in \mathbb{N}$ with $m,n > N$. Obviously, by Property 1, we have both $|x_m - x| < \varepsilon / 2$ and $|x_n - x| < \varepsilon / 2$. Therefore, we have \begin{equation*} \begin{split} |x_m - x_n| & = |x_m - x_n + 0| \\ & = |x_m - x_n + (x - x)| \\ & = |(x_m - x) + ( - x_n + x)| \\ & \le |x_m - x| + | - x_n + x| && | \text{ by subadditivity of abs. val.} \\ & = |x_m - x| + |x_n - x| && | \text{ by evenness of abs. val.} \\ & < \varepsilon / 2 + \varepsilon / 2 \\ & = \varepsilon. \end{split} \end{equation*} Thus, by transitivity, $|x_m - x_n| < \varepsilon$. QED