If $|x_{n+1} - x_n| < \epsilon,$ is $(x_n) $ a Cauchy sequence?

I disagreed that this proof was valid, because the sequence $x_n := \sqrt{n} \, $ doesn't satisfy the hypothesis, particularly that if $\forall \epsilon > 0$

$$ | x_{n+1} - x_n| < \epsilon\, , \quad \forall n \geq N$$ is satisfied, then the hypothesis requires that
$$ | x_{n+2} - x_{n+1}| < \epsilon\, , \quad $$ is also satisfied, and so on.

Yes, and all the inequalities you list are satisfied. In particular, since

$$|\sqrt{n+1} - \sqrt{n}| = \frac{1}{\sqrt{n+1}+\sqrt{n}}\leq \frac{1}{\sqrt{n}}$$

you have, for every $\epsilon > 0$, the selection $N=\lceil\frac{1}{\epsilon^2}\rceil$ to prove it. From this alone, we can already be certain that your proof is incorrect (because $x_n=\sqrt{n}$ is obviously a good counterexample).

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To see where your proof went wrong, you correctly prove that $|x_m-x_n|<(m-n)\epsilon$, however to prove that $x_n$ is cauchy, you should be able to prove that $|x_m-x_n|$ is arbitrarily small, but the expression contains the value $(m-n)$ which you were not able to bound. Yes, you could have every term in a sum bounded by one trillionth, but there is nothing to guarantee that $m-n$ is smaller than one quadrillion. The problem is you are given $\epsilon$ in advance, and then the inequality has to hold for every pair $m,n$.