In an attempt to prove Theorem 1.2 below, I shall first establish Theorem 1.0 and Theorem 1.1, the Archimedean Property of the Real Number System.
Let Theorem 1.0 be that the set of positive integers, $\mathbb Z^+$, is unbounded above. This may be proved via the Completeness Axiom, the details of which are left to the reader.
As a corollary to Theorem 1.0, we may establish Theorem 1.1 as follows.
Theorem 1.1: $\forall x, y$ where $ x > 0$ and $ y \in \mathbb R$, $\exists n \in \mathbb Z^+$ such that $nx > y$.
Proof. Algebraic manipulation of $nx > y$ yields $n > \frac{y}{x}$, so if Theorem 1.1 was false, $\frac{y}{x}$ would be an upper bound for the set $\mathbb Z^+$ of positive integers, contradicting Theorem 1.0.
Theorem 1.2 If three real numbers $a$, $x$, and $y$ satisfy the inequalites $$a \le x \le a + \frac{y}{n}$$ $\forall n \in \mathbb Z^+$, then $x = a$.
Proof. We shall break this proof into two sections. First, assume, for the purposes of contradiction, that $x > a$. This implies that $x - a > 0$, so by Theorem 1.1, we may say that that there exists some positive integer $n$ that satisfies the inequality $n(x - a) > y$. Conversely, we may assume, for the purposes of contradiction, that $x < a$, in which case $a - x > 0$, and we can again say that there exists some positive integer $n$ satisfying the inequality $n(a - x) > y$. However, simple algebraic manipulation of the original inequality of the theorem in question yields $$n(x - a) \le y.$$ It's clear to see, therefore, that neither the result $n(x - a) > y$ nor $n(a - x) > y$ from our previous assumptions was correct, leading to a contradiction. If $x \ngtr a$ and $x \nless a$, by the Law of Trichotomy, $x$ must equal $a$. $\square$
I am wondering if my proof is correct, and whether or not there would be any better ways to write it. Thank you.
Yes your proof is correct. You can make it lighter and simpler by not treating the case $x<a$ since the hypothesis is the contrary.