Proof Verification: A monotonically increasing sequence that is bounded above always has a LUB 
Problem: Prove that a monotonically increasing sequence that is bounded above always has a least upper bound.

This theorem is present in calculus and real analysis books along with proofs for it. I want to know if my proof is convincing enough.
Solution: let's start with the definitions

If $ \{a_n\} $ is monotonically increasing, then $ a_{n+1} \geq a_n $ for all $n$.
If $ \{a_n\} $ is bounded above then there exists $M$ such that $a_n \leq M$ for all $n$.
If $ M $ is an upper bound, but no number less than $ M $ is an upper bound for $ \{a_n\} $, then $ M $ is called the least upper bound for $ \{a_n\}$.

It has already been established in the previous exercise that if a monotonically increasing sequence $ \{a_n\} $ is bounded above, then the sequence converges to some value. Let's call this value of convergence $\ell$. Then
$$ lub\{a_n\} = max\{a_n\} = \ell = \lim_{n \to \infty} a_n $$
For if $ \ell $ wasn't the least upper bound of $ \{a_n\} $ then $\{a_n\}$ wouldn't converge to it.
 A: It's not quite convincing enough for me. You didn't explicitly show why it was true. Let $L$ be the limit of the sequence $\{a_n\}$ We'll use the monotone convergence theorem as you said it was already proven to you. Clearly $L \geq a_n$ for all $n$. And so $L$ is an upper bound. 
Assume there is some upper bound which is less than $L$, call it $k$. Let $L - k = p$. From the definition of the limit of a sequence, there exists an $N$ such that for all $n > N$, $|a_n - L| < \varepsilon$ for all $\varepsilon > 0.$ Now take $\varepsilon = p$. This shows that there is a term which is greater than $k$, contradicting the choice of $k$ as an upper bound. 
Now we have shown  that $L$ is a upper bound, and any value smaller than $L$ is not an upper bound, finishing the proof.
A: My Approach-
Let M be the LUB of {$x_n$}.
Then $x_n \in N_ε(M)$ for some $n\in S\forall  ε>0$ ,(i. e.$x_n \in N_ε(M)$ when $K\leq n\leq s^ * $ , where $K:=inf S , s^ *:=sup S$).Here S is an interval in $\Bbb R$ ,such that $infS,supS\in\Bbb N$.
But let $\exists k>s^*=supS$ such that $x_k \notin N_ε(M).(\therefore k>n)$
$\implies x_n>x_k$,viz. a contradiction. $^. .^. (x_n)$ is monotonically increasing.
$\therefore \nexists k\in\Bbb N :k>s^*$ for which $x_k \notin N_ε(M) \forall ε>0$.
$\implies \forall ε>0 ,\exists K $ such that $x_n\in N_ε(M) \forall n\geq K.$
Hence {$x_n$}$\to M$. ■
