Let $(x_n)_{n\in\Bbb N}$ be a real sequence.
$\textbf{Definition 1.}$ $(x_n)$ is $\textit{convergent}$ iff there is an $x\in\Bbb R$ such that, for every $\varepsilon\in\Bbb R$ with $\varepsilon>0$, there is an $N\in\Bbb N$ such that, for every $n\in\Bbb N$ with $n>N$, we have $|x_n-x|<\varepsilon$.
$\textbf{Definition 2.}$ $(x_n)$ is a $\textit{Cauchy sequence}$ iff, for every $\varepsilon \in\Bbb R$ with $\varepsilon > 0$, there is an $N\in\Bbb N$ such that, for every $m,n\in\Bbb N$ with $m,n > N$, we have $|x_m - x_n| < \varepsilon$.
$\textbf{Theorem.}$ If $(x_n)$ is convergent, then it is a Cauchy sequence.
Proof: Since $(x_n)\to x$ we have the following for for some $\varepsilon_1, \varepsilon_2 > 0$ there exists $N_1, N_2 \in \Bbb N$ such for all $n_1>N_1$ and $n_2>N_2$ following holds $$|x_{n_1}-x|<\varepsilon_1\\ |x_{n_2}-x|<\varepsilon_2$$ So both will hold for all $n_1, n_2 >\max(N_1, N_2)=N$, say $\varepsilon = \max(\varepsilon_1, \varepsilon_2)$ then $$|x_{n_1}-x-(x_{n_2}-x)|<\varepsilon\\\implies |x_{n_1}-x_{n_2}|<\varepsilon$$ Hence all convergent sequences are Cauchy.
Is this proof correct? I also saw this question and copied some of the content(definition and theorem) from there.https://math.stackexchange.com/q/1105255