Preliminary Note
Though this question has been asked before, I haven't yet found a solution on the site that shows the existence of such an integer using only Cauchy sequences and their properties. If this is only because I have not searched hard enough, please label me as a duplicate and let me know where. This question is a proof verification.
Note that the properties of the real number line concerning least upper bounds and completeness have not been explained yet in this textbook (Tao Analysis 1), and all we have done at this point is construct the real number system and give it definitions of order and arithmetic that allow it to be a field. Also, note that we've already proved a similar property with the rational (just replace real below with rational numbers).
Question
Show that for every real number $x$ there is exactly one integer $M$ such that $M \leq x < M+1$.
Attempt
This outlines my attempt at a proof and might highlight some definitions or idiosyncrasies that my textbook has.
Let $x$ be a real number. Then $x = \lim_{n \rightarrow \infty} (a_n)_{n=1}^{\infty}$ for some Cauchy sequence $(a_n)_{n=1}^{\infty}$. Since $(a_n)_{n=1}^{\infty}$ is Cauchy, we know that for all $\epsilon > 0$, there exists a $N \geq 1$ such that for $j,k \geq N$, $|a_j - a_k| \leq \epsilon$. Consider some $\epsilon' > 0$ and its corresponding $N$. Take $a_N$. Since $N \geq N$, $|a_j - a_N| \leq \epsilon'$ for all $j \geq N$. Now let us construct a new sequence $(b_n)_{n=1}^{\infty}$ where $b_n = a_N$ if $n \leq N$ and $b_n = a_n$ if $n > N$. It is clear that $x = \lim_{n \rightarrow \infty} (b_n)_{n=1}^{\infty}$ as well. Now note that since $|a_j - a_N| \leq \epsilon'$, $- \epsilon \leq a_j - a_N \leq \epsilon$. So $a_N - \epsilon' \leq a_j \leq a_N + \epsilon'$. By a previous proposition, we already know that there exists some integer $M$ such that $M \leq a_N < M+1$. Since $M \leq a_N$, we have two cases to consider. If $M < a_N$ simply choose an $\epsilon'$ small enough such that $a_N - \epsilon' \geq M$ and $a_N + \epsilon' < M+1$. Then since every value of $(b_n)_{n=1}^{\infty}$ is such that $M \leq b_n < M+1$, we know that $M \leq x \leq M+1$. If $x = M+1$, let the integer be $M+1$. Then $M+1 \leq x < M+2$. So there always exists some integer $M$ such that $M \leq x < M+1$. But if we have the special case that $M = a_N$ no matter how big the $N$ and no matter how small an $\epsilon$ we choose, we can then conclude that $x = M$, so $M \leq x < M+1$. The uniqueness of this integer is a simple proof and is found elsewhere on this website.
The proof that this $\epsilon'$ exists is as follows. Note that $a_N$ (by trichotomy) must be greater than, equal, or less than $M + 1/2$. If $a_N > M+1/2$, take $\epsilon' = (M+1-a_N)/2$. Then since $a_N < M+1$ by definition, $2a_n + M+1-a_N < 2M+2$ and $a_n + (M+1-a_N)/2 < M+1$. Similarly, since $a_N > M+1/2$, we can conclude that , $3a_N \geq 3M + 3/2 > 3M+1$, and so $2a_N -M - 1 + a_N \geq 2M$ and $a_N - (M+1-a_N)/2 \geq M$. Without loss of generality, if $a_N < M+1/2$, take $\epsilon' = (a_N - M)/2$, and if $a_N = M+1/2$, take any $\epsilon' < 1/2$. This concludes our proof of the existence of such an $\epsilon'$.