Here's my attempt:
Lemma: If a real number $x = \lim_{n\to \infty} x_n > 0$, then the terms of the Cauchy sequence $\langle x_n \rangle$ must be eventually positive, i.e., there exists $N \in \mathbb{N}$ s.t., $\forall n\geq N$, $x_n >0$.
Proof: By definition of $x>0$, $x = \lim_{n\to \infty} y_n$ for some $\langle y_n \rangle $ that is positively bounded away from zero, so that $\forall n\geq N, y_n \geq w$, where $w>0 \in \mathbb{Q}$. Since $\lim_{n\to \infty} x_n = \lim_{n\to \infty} y_n$, by definition $\langle x_n \rangle = \langle y_n \rangle$. Then if we pick $\epsilon = \frac{w}{2}>0$, there exists $N$ s.t. $\forall n \geq N, |x_n - y_n| \leq \frac{w}{2}$, which implies $x_n - y_n \geq -\frac{w}{2}$; adding $y_n \geq w$ to both sides gives $x_n \geq \frac{w}{2} > 0$.
On to the real proof:
Since $a > c$, the formal limit of the Cauchy sequence of rationals $\langle a_n - c\rangle$ must be positive. By definition, this means $\lim_{n\to \infty} a_n -c = \lim_{n\to \infty} b_n$ for some Cauchy sequence $\langle b_n \rangle$ that is positively bounded away from zero, i.e., $\exists d >0, \forall n, b_n \geq d$. By definition of equality of real numbers, the sequences $\langle a_n -c \rangle$ and $\langle b_n \rangle$ are equivalent, so for every $\epsilon >0$, there exists $N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $|a_n - c - b_n| \leq \epsilon$.
Since $a>0$, by our lemma there exists $N_2$ such that $\forall n\geq N_2, a_n > 0$. Let $N = \max \{N_1, N_2\}$.
Now assume for contradiction that it's not the case that eventually $a_n > c$; then there must exist $n_0\geq N$ with $a_{n_0} \leq c$, for otherwise eventually $\forall n\geq N, a_n > c$, a contradiction.
Then we have
$$c + d - a_{n_0} \leq c + b_n- a_{n_0} = |c + b_n| - |a_{n_0}| \leq |a_{n_0} - c - b_n| \leq \epsilon $$ so $a_{n_0} \geq c + d - \epsilon$.
If we choose $\epsilon=d/2$, then $a_{n_0} \geq c + d/2$, contradicting our assumption that $a_{n_0} \leq c$.
Is this correct? Any simpler or alternative proof is also appreciated.
