Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers which has some finite upper bound $M\in\textbf{R}$, and which is also increasing (i.e., $a_{n+1}\geq a_{n}$ for all $n \geq m$). Then $(a_{n})_{n=m}^{\infty}$ is convergent, and in fact \begin{align*} \lim_{n\rightarrow\infty}a_{n} = \sup(a_{n})_{n=m}^{\infty} \leq M \end{align*}
MY ATTEMPT
Let $s = \sup(a_{n})_{n=m}^{\infty}$. We want to prove that, for any positive real number $\varepsilon > 0$, there exists a natural number $N\geq m$ such that $|a_{n} - s|\leq\varepsilon$ whenever $n\geq N$, which is equivalent to $s-\varepsilon < a_{n} < s + \varepsilon$ whenever $n\geq N$.
Indeed, according to the properties of the supremum, for every $\varepsilon > 0$, there exists a $n_{0}$ such that $s - \varepsilon < a_{n_{0}}\leq s$.
Since $a_{n}$ is increasing, $s - \varepsilon < a_{n_{0}} < a_{n} \leq s < s + \varepsilon$ whenever $n > n_{0}$.
Consequently, we have proven that, for any $\varepsilon > 0$, there is a natural number $n_{0}\geq m$ such that
$$|a_{n} - s|\leq\varepsilon$$
whenever $n\geq n_{0}$, and we are done.
Can someone double-check my argument?
As Mark indicated in a comment, and I also confirm, your argument is correct.