In some earlier work, I proved that any non-empty subset of $\mathbb{R}$ that is bounded below has a greatest lower bound.
I now want to prove that for any $x\in\mathbb{R}$, there exists $k\in\mathbb{Z}$ such that $k\leq x$.
Attempted proof
Fix $x\in\mathbb{R}$ and suppose that no such integer $k$ exists.
Then for all $a\in\mathbb{Z}$, we have that $a>x$. Thus, $x$ is a lower bound for $\mathbb{Z}$.
By the theorem I proved earlier, $\mathbb{Z}$ therefore has a greatest lower bound. Let's call it $\alpha$.
This means that $\alpha+1/2$ is not a lower bound, and thus, there exists some $b\in\mathbb{Z}$ such that $\alpha<b<\alpha+1/2$.
This means that $b-1$ is an integer less than $\alpha$, and hence, we have a contradiction.
Is this proof sufficient?