Cauchy completeness plus the Archimedean property imply the monotone convergence property

Suppose $(x_k)$ is not Cauchy in $F$: there is $\varepsilon > 0$ in $F$ such that for every $N \in \mathbb{N}$ there are $k, l \ge N$ with $|x_k - x_l| \ge \varepsilon$.

For every $n \in \mathbb{N}$ there is $k > n$ with $x_k - x_n \ge \varepsilon$: apply step 1.1 with $N := n + 1$ to get $k, l \ge n+1$ with $|x_k - x_l| \ge \varepsilon$, name them so that $l \le k$, note that monotonicity gives $x_l \le x_k$ and hence $|x_k - x_l| = x_k - x_l \ge \varepsilon$, and note that $n < l$ gives $x_n \le x_l$, so $x_k - x_n \ge x_k - x_l \ge \varepsilon$ with $k \ge n+1 > n$.

For each $n$ the set $\{\, k \in \mathbb{N} : k > n \text{ and } x_k - x_n \ge \varepsilon \,\}$ is therefore nonempty and has a least element, so $f(n) := \min\{k : k > n, \ x_k - x_n \ge \varepsilon\}$ is a total function $\mathbb{N} \to \mathbb{N}$; the recursion theorem applied to $\mathbb{N}$, the element $0$ and $f$ gives indices $k_0 = 0$ and $k_{j+1} = f(k_j)$, with $k_j < k_{j+1}$ and $x_{k_{j+1}} - x_{k_j} \ge \varepsilon$ for every $j$.

By induction on $j$, $x_{k_j} - x_{k_0} \ge (j \cdot 1_F)\,\varepsilon$ for every $j$: at $j = 0$ both sides are $0$, and adding $x_{k_{j+1}} - x_{k_j} \ge \varepsilon$ to the inductive inequality gives $x_{k_{j+1}} - x_{k_0} \ge (j \cdot 1_F)\varepsilon + \varepsilon = ((j+1)\cdot 1_F)\,\varepsilon$.

Since $\varepsilon > 0$ is invertible with $\varepsilon^{-1} > 0$, the Archimedean property supplies $j \ge 1$ with $(B - x_{k_0})\varepsilon^{-1} < j \cdot 1_F$, hence $B - x_{k_0} < (j \cdot 1_F)\varepsilon$ and $x_{k_j} \ge x_{k_0} + (j \cdot 1_F)\varepsilon > B$, contradicting the hypothesis that $B$ bounds every term of $(x_k)$.

The assumption of step 1.1 is therefore untenable, so $(x_k)$ is Cauchy in $F$ and (CC) makes it converge in $F$; as $(x_k)$ was an arbitrary nondecreasing sequence bounded above, $F$ has (MCT).