While your equation is correct as stated, $m$ can be arbitrarily large, which can make the factor $m - n - 1$ preceding your $\epsilon$ arbitrarily large, in turn making $(m - n - 1) \epsilon$ arbitrarily large.
$a_n := \sum_{i = 1}^n \frac{1}{i}$ is a sequence obeying the alternative property you've given, but it is not convergent.
Your proof is correct. Intuitively, it should make sense--if you have a sequence that stays a bounded distance above $0$, then any other sequence getting close to it will also say above $0$. It doesn't matter whether your first sequence was actually closing in on any fixed positive number, if all you care about is staying above $0$.
Note, though, that the characterization of positiveness which you use from Theorem 1 does use the fact that $(a_n)$ is Cauchy--the original definition of positiveness is weaker for general sequences, and Theorem 3 is not true for general sequences with the original definition of positiviteness. Roughly, you could have a sequence which oscillates between large positive values and positive values approaching $0$, and then a sequence could be co-Cauchy with it while being negative on the part that approaches $0$, so it does not have a positive tail. There are a couple different natural ways you might define a positive general sequence so that Theorem 3 remains true--you could either take the condition of Theorem 1 as the definition, or you could weaken the positive tail condition to say that for any rational $\epsilon>0$, $(a_n+\epsilon)$ has a positive tail.
The real place where the assumption that the sequences are Cauchy will be crucial for the theory of the ordering of real numbers is that this notion of "positivity" gives a total order. In other words, every Cauchy sequence is either positive, converges to $0$, or is negative (meaning if you negate its terms you get a positive sequence). For arbitrary sequences you could have sequences that oscillate between positive and negative values and so are neither positive nor negative. (Relatedly, you also need the Cauchy condition in order to get multiplicative inverses, since you need any sequence which does not converge to $0$ to stay away from $0$ in its tail.)
(Or, in the language of abstract algebra, if you take the set of all sequences modulo the co-Cauchy equivalence relation, you will still get a partially ordered commutative ring. However, you will not get a totally ordered field which is what you want the real numbers to be.)
Best Answer
Yes, that would be correct. In other words: if $(a_n)_{n\in\mathbb N}$ is a Cauchy sequence which is convergent in $\mathbb R$ (note that this is equivalent to the assertion that it is a Cauchy sequence), then its limit $l$, seen as Dedekind cut is precisely the set$$\{x\in\mathbb Q\,|\,(\exists n\in\mathbb N)(\forall m\in\mathbb N):m\geqslant n\implies x<a_m\}$$(I used $\leqslant$ here instead of $<$ just for a matter of taste).