Is the following argument correct?
Proposition. If $\{x_n\}$ is Cauchy sequence such that $x_n = c$ for infinitely many $n$, then $\lim_{n\to\infty}x_n = c$.
Proof. Let $\epsilon>0$. Since $\{x_n\}$ is a Cauchy sequence, there exists an $M\in\mathbb{N}$ such that $\forall \, j,k\ge M$, we have $|x_j-x_k|<\epsilon$. Now, since $x_n = c$ for infinitely many $c$, then surely $x_r = c$ for some $r \ge M,$ implying that $|x_j-c|<\epsilon \,\,\forall j\ge M$, completing the argument.
$\blacksquare$
Best Answer
As told in the comments, your proof is fine.
Usually when I am asked to prove a statement that seems rather obvious, I tend to prove it by contradiction: this usually allows me to explore the intuitive feel I have for the result.
Assume the sequence converged to $l \neq c$. Then for $\epsilon = \frac{|c - l|}{2} $ $\exists N $ s.t. $i > N \implies |x_i - l| < \epsilon $, by definition of convergence. But this is absurd, because the sequence had infinitely many terms equal to $c $ and thus infinitely many $x_k = c $ with $k > N $.