$\forall x \in \mathbb{R}^+ ( \exists M \in \mathbb{Z}^+ ( x > 1/M > 0))$: Cauchy Sequences Synopsis
I'm sure this question can be found elsewhere on this website, but I haven't yet found a perspective though that uses the construction of the reals through Cauchy sequences to prove this statement. The reason why I'm using Cauchy sequences to prove this is because my textbook (Tao Analysis 1) builds up the reals first before delving into the more well known properties of it (like the Axiom of Completeness). I would like to ask the community here just to help verify the validity of my proof.
Note that the definitions Tao uses are often non-standard and peculiar. So in my proof, I'll try my best to highlight any definitions I use and expand them out. Also note that we've already proved that for every rational $q$, there exists an integer $N$ such that $N \leq q < N+1$.
Exercise

Show that for any positive real number $x>0$ there exists a positive integer $N$ such that $x > 1/N > 0$.

Proof
Since $x$ is a positive real number, $x = \lim_{n \rightarrow \infty}a_n$ for some rational sequence $(a_n)_{n=1}^{\infty}$ positively bounded away from zero. In other words, there exists a rational $c > 0$ such that $a_n \geq c$ for all $n \geq 1$. Because $(a_n)_{n=1}^{\infty}$ is Cauchy, we know that for each $\epsilon > 0$, there exists some $N \geq 1$ such that for each $j,k \geq N$, $|a_j - a_k| \leq \epsilon$. Now take $\epsilon = c/2$ and its corresponding $N$. Then $a_N - \epsilon > 0$. Now construct a new sequence $(b_n)_{n=1}^{\infty}$ such that if $n \leq N$, then $b_n = a_N$, and if $n > N$, then $b_n = a_n$. Since for all $j,k > N$, $-\epsilon \leq a_j - a_k \leq \epsilon$, it is clear that for all $n \geq 1$, $a_N - \epsilon \leq b_n \leq a_N + \epsilon$. Now we wish to find some integer $M > 0$ such that $1/M < a_N - \epsilon$. So we must find some integer $M$ such that $M > 1/(a_N - \epsilon)$. But we know that such an integer exists by a previous proposition, and since $1/M < a_N - \epsilon \leq b_n$ for all $n \geq 1$, we can conclude that $0 < 1/M \leq x$. If $1/M < x$, then the exercise is proven. If $1/M =x$, just take $M+1$. Then the exercise is satisfied. So either way, we have found a positive integer $M$ such that $x > 1/N > 0$.
 A: You can make it clearer by specifying that $x$ is the limit of a rational sequence. You also need a bit more detail about how that sequence is bounded away from $0$: there are an $\epsilon>0$ and an $n_0\in\Bbb Z^+$ such that $a_n\ge 2\epsilon$ for each $n\ge n_0$. Now use the fact that the sequence is Cauchy to say that there is an $n_1\ge n_0$ such that $|a_j-a_k|\le\epsilon$ whenever $j,k\ge n_1$: this ensures that $a_{n_1}-\epsilon>0$, since $a_{n_1}\ge 2\epsilon$.
Now define the $b_n$ as before: $b_n=a_{n_1}$ if $n\le n_1$, and $b_n=a_n$ otherwise. Your argument that $a_{n_1}-\epsilon\le b_n\le a_{n_1}+\epsilon$ for all $n\ge 1$ is fine, but after that you have a problem: all of a sudden you’re using $N$ for two different integers, my $n_1$ and a new one that I’m going to call $m$. What you want now is some positive integer $m$ such that $\frac1m<a_{n_1}-\epsilon$ or, equivalently, $m>\frac1{a_{n_1}-\epsilon}$. Then you can argue that $\frac1m<a_{n_1}-\epsilon\le b_n$ for all $n\ge 1$ and conclude that $0<\frac1m\le x$. At that point there’s really no need to split the argument into cases depending on whether that last inequality is strict: you might as well just observe that $0<\frac1{m+1}<\frac1m\le x$.
