Here's my best guess. Dedekind:
Hmmmmmm.
The rational number line is full of holes. At $\sqrt{2}$, there should be a number, but there isn't.
What does that even mean?
Hm.
Okay, there is no hole at $2$. We can make this precise by observing that $(2-\epsilon,2)_\mathbb{Q}$ is distinct from $(2-\epsilon,2]_\mathbb{Q},$ for all strictly positive $\epsilon$.
But there is a hole at $\sqrt{2}.$ We can make this more precise by observing that $(\sqrt{2}-\epsilon,\sqrt{2})_\mathbb{Q}$ has exactly the same elements as $(\sqrt{2}-\epsilon,\sqrt{2}]_\mathbb{Q}$ for all strictly positive $\epsilon$.
But wait. If we're trying to build the real numbers, then the notation $\sqrt{2}$ isn't 'allowed' yet. Okay, so lets make it precise like this.
There is a hole at $\sqrt{2},$ which means that $\{x \in \mathbb{Q} \mid x^2 < 2\}$ has exactly the same elements as $\{x \in \mathbb{Q} \mid x^2 \leq 2\}.$
But no, this is too symmetrical; the above statement could equally well be seen as the claim that there's a hole at $-\sqrt{2}$. Okay, lets go ahead and destroy that symmetry.
There is a hole at $\sqrt{2},$ which means that the downward closure of $\{x \in \mathbb{Q} \mid x^2 < 2\}$ has exactly the same elements as the downward closure of $\{x \in \mathbb{Q} \mid x^2 \leq 2\}.$
Aha! So maybe we should define that $\sqrt{2}$ is the downward closure of $\{x \in \mathbb{Q} \mid x^2 < 2\}.$ But wait, why not the downward closure of the non-strict version? Okay, lets just disregard that possibility for the moment.
So, a real number is the downward closure of a set of rational numbers that is bounded above, but has no greatest element.
Wait. How about (Eureka!):
A real number is a downward-closed set of rational numbers that is bounded above, but has no greatest element.