Prove if $\{s_n\}$ is monotonically increasing. Then $\{s_n\}$ converges if and only if it is bounded.
My Proof
Since $\{s_n\}$ is monotonically increasing, $s_n \leq s_{n+1}$.
Part 1: We first assume $s_n \to p$, and show that $\{s_n\}$ is bounded.
Since $s_n \to p$ and $s_n \leq s_{n+1}$, $p$ is an upper bound of $\{s_n\}$, as $p \geq s_n$ for all $n$. To show that it is the least upper bound, take $\epsilon > 0$, then $p - \epsilon < p$ is not an upper bound of $\{s_n\}$ and $p = \sup \{s_n\}$
Now we also have $s_0$ as a lower bound of $\{s_n\}$, as $s_0 \leq s_n$ for every $n$. To show that $s_0$ is the greatest lower bound, we fix $\epsilon > 0$, then $s_0 + \epsilon > s_0$ is not a lower bound of $\{s_n\}$. Therefore $s_0 = \inf \{s_0\}$. And thus $\{s_n\}$ is bounded.
Part 2: Conversely we assume $\{s_n\}$ is bounded and use it to show $s_n \to p$.
$p = \sup \{s_n\}$ and $s_0 = \inf\{s_n\}$ by the same arguments made above. And since $s_n \leq s_{n+1}$ and $p = \sup \{s_n\}$, it follows that $s_n \to p$.
Thus a monotonically increasing $\{s_n\}$ converges if and only if it is bounded. $\square$
Is my proof correct, and if so how rigorous is it? I'm specifically looking for comments on my level of rigor, I don't want my proofs to be hand-wavy or sketchy at all. Also any comments on my proof-writing is greatly appreciated.