Is the Following Proof Correct?
Proposition. Let $a, b$ be rational numbers. Show that $a = b$ if and only if $\text{LIM}_{n\to\infty}a = \text{LIM}_{n\to\infty}b$ (i.e., the Cauchy sequences $a, a, a, a,\dots$ and $b, b, b, b,\dots$ are equivalent if and only if $a = b$). This allows us to embed the rational numbers inside the real numbers in a well-defined manner.
Proof. Assume $a=b$ and let $\epsilon>0$ and $N$ be any natural number, evidently given any $n\ge N$ we have $|a_n-b_n| = |a-b| = 0\leq\epsilon$ implying that $(a)_{n=0}^{\infty}$ and $(b)_{n=0}^{\infty}$ are equivalent sequences.
Conversely let $(a)_{n=0}^{\infty}$ and $ (b)_{n=0}^{\infty}$ be equivalent sequences consequently given any $\epsilon>0$ for some $N\in\mathbf{N}$ we have $\forall n\ge N(|a_n-b_n| = |a-b|\leq\epsilon)$. Proposition $\textbf{4.2.9}$ establishes that exactly one of $a<b$, $a>b$ or $a=b$ may be true.
Assume that $a<b$ consequently $(b-a) = x\in\mathbf{Q}^+$ then by hypothesis $|a_N-b_N| = |b_N-a_N| = |x|\leq\frac{|x|}{2}$ a contradiction. We may infer a similar contradiction by assuming $a>b$ consequently it must be that $a=b$.
$\blacksquare$
YES, your proof is correct.
And here's an easier way to prove the statement:
$\lim_{n \to \infty} a = \lim_{n \to \infty} b$
$\implies \lim_{n \to \infty} (a \cdot 1) = \lim_{n \to \infty} (b \cdot 1)$
$\implies a \cdot \lim_{n \to \infty} 1 = b \cdot \lim_{n \to \infty} 1$
$\implies a \cdot 1 = b \cdot 1$
$\implies a=b.$