I'm having some trouble proving that BW implies MCT. Here's what I've done so far
If a bounded sequence $(a_n)$ is monotone, then the sequence is convergent.
Case 1: $(a_n)$ is always increasing.
Proof. By BW, there exists a subsequence $(a_{n_k})$ that converges to some limit $L$. Observe that $$|a_n-L| = |a_n-a_{n_k}+a_{n_k}-L|\leq |a_n-a_{n_k}|+|a_{n_k}-L|$$ I think the next step is to argue that you can make the left term as small as possible but I'm not sure how to show that.
On
You are close.
MCT : Every bounded monotone sequence is convergent.
BW: There exist a convergent subsequence $(a_{n_k})$.
Consider a monotonically increasing sequence.
Let $\epsilon >0$ be given.
There is a $k_0$ such that for $k \ge k_0$:
$|a_{n_k} -L|< \epsilon $.
Let $N:= n_{k_0}$.
For every $n \ge N$ there is a $k \ge k_0$ with
$n_k \le n \lt n_{k+1}$.
Since $(a_n)$ is mon. increasing we have
$a_{n_k} \le a_n \le a_{n_{k+1}} \le L$,
hence
$|a_n-L| \le |a_{n_k} -L| \lt \epsilon$.
Note that $L>a_{n}$ for all n (why?). Since $a_{n_{k}}$ converges to $L$, $L-a_{n_{k}}$ is smaller than a given $\varepsilon$ if k is bigger than some Index $K$. But now since $a_{n}$ is increasing, $L-a_{n}$ must also be smaller than $\varepsilon$ for $n>n_{K}$.