I have a vague idea of what to do for the first part, which I've sketched below but I'm confused about how to prove the condition (1).
Thank you in advance for taking the time to read this.
So if we have this set
$$S = \{max(m\in\mathbb{Z}): m\leq x, x\in\mathbb{Q}\}$$
We know its non-empty because we can just pick some $x$, say $x=2.7$, and there exists $m=2$ in the set $S$ which satisfies the condition.
We know $x$ is an upper bound for $S$ because for every $m\in S$ there is some $x$ greater than or equal to it.
Then by the Least Upper Bound Property, $S$ has a supremum, call it $r$.
By Lemma 1.2, for some $t<r$ there exists some $m\in S$ such that $m>t$. So if we set $t=r-1$, then there exists some $m\in S$ such that $m>r-1$. By definition, $m\leq x$.
So we have $m\leq x$ and $m>r-1$, which looks similar enough to (1) but I'm unsure how I'm supposed to get $x$ instead of $r$ in this process. I would greatly appreciate some guidance on this.