Here's an attempt at a proof:
Proposition: Completeness Axiom $\iff$ Monotone Convergence Theorem
Proof:
$\Leftarrow$
Assume for contradiction that Completeness Axiom $\not\Leftarrow$ MCT. That is, $x_n$ is monotone increasing and bounded sequence of real numbers with a finite limit. By our assumption, $\displaystyle\lim_{n \to \infty} x_n = u > \text{sup }x_n$. However, this is a direct contradiction of the Monotone Convergence Theorem since a monotone sequence converges to sup $a_k$ if $a_k$ is increasing. Therefore, the Completeness axiom $\Leftarrow$ Monotone Convergence Theorem
$\Rightarrow$
Suppose that the sequence $x_n$ is monotone increasing and bounded. Then $x_n$ is bounded, so it has a least upper bound $u \equiv \text{sup }x_n$ by the completeness axiom. Now for every $\epsilon > 0$, there exists an $N$ such that $N > u - \epsilon$, since otherwise $u- \epsilon$ is an upper bound of $x_n$, which contradicts to $u$ being the supremum $x_n$. Then, since $x_n$ is increasing, for all $n>N$ we have $| c - x_n | = c - x_n \leq c - x_N < \epsilon$. Hence by definition, the limit of $x_n = u$. Therefore, the Completeness axiom $\Rightarrow$ Monotone Convergence Theorem