$\mathbb{R}$ is a set. By one construction, its elements are precisely those sets called "cuts".
$\mathbb{Q}$ is a set. None of its elements are "cuts."
Thus, $\mathbb{Q} \cap \mathbb{R}$ must intersect on the empty set. However, most mathematicians would find such logic overly pedantic (I think).
How can I reconcile this?
On
Well, sure, I suppose in a strictly set-theoretic sense you're right. But I think this ignores the pragmatic, human aspect of mathematics.
It only really makes sense to talk about the intersection of two sets if they somehow "live in the same world". For instance, you could define $\mathbb{R}$ as a set of Dedekind cuts, or you could define $\mathbb{R}$ as a set of decimal expansions (under some appropriate equivalence relation), and both would be perfectly good definitions of $\mathbb{R}$; but Dedekind cuts and decimal expansions aren't the same thing, so does that mean that $\mathbb{R}$ (the first one) $\cap \mathbb{R}$ (the second one) $= \varnothing$? I suppose so, yes. But this isn't a profound mathematical fact so much as crappy notation: these don't both deserve to be called $\mathbb{R}$, because they have nothing to do with each other yet.
In reality, the set of Dedekind cuts is one nice way of modelling $\mathbb{R}$, and the set of decimal expansions is another. Neither of them is $\mathbb{R}$ any more than the other is. You can define one to be your favourite, canonical copy of $\mathbb{R}$ if you like, but then suddenly the other isn't $\mathbb{R}$ any more. In order to meaningfully call them both $\mathbb{R}$, you need to also specify some way of identifying their elements.
Likewise, when you construct $\mathbb{R}$ from $\mathbb{Q}$ as a set of Dedekind cuts, if you try to take the naive set-theoretic intersection then you get something trivial, of course. But this ignores the fact that we want to construct $\mathbb{Q}$ as something sitting inside $\mathbb{R}$ - otherwise, why are we doing it in the first place, if we now have two different and completely incomparable notions of the number $5$?
This is why set-theoretic equality is usually way too strong a notion for most mathematical purposes, and we often deal with things like isomorphisms, natural isomorphisms, equivalences under various relations, etc.
On
Consider this picture:
"Universe of numbers": $\ \mathbb Q \subset ??????$
mapping between universes: $\updownarrow\updownarrow$
"Universe of dedekind cuts": $\mathbb Q^* \subset \mathbb R^*$
$\mathbb Q$ are the rational numbers. It's an ordered field. In particular it is an ordered field generated by $\langle 1 \rangle$.
$\mathbb R^*$ is the set of all dedekind cuts of the rational numbers. If $A\le B$ is defined as $A \subset B$ then $\mathbb R^*$ is an ordered field.
Now $\mathbb Q^*$ is a subset of $\mathbb R^*$. It is the subset of all cuts which actually have a rational least upper bound (as opposed to the cuts of which there is no rational least upper bound). This is a subfield of $\mathbb R^*$.
And $\mathbb Q^*$ is equivalent in every sense of the word to the field $\mathbb Q$.
So there is an equivalence mapping between the "two universes".
But that means if we "retro-equivalence" back from $\mathbb R^*$ in the universe o cuts, back to the universe of numbers, we co retroactively discover an ordered field $??????$ that is equivalent to $\mathbb R^*$ and of which $\mathbb Q$ is a subfield.
That field is $\mathbb R$, the ordered field of the real numbers, in which every real number is the least upper bound of a dedekind cut.
....
Okay, to put it in a less esoteric woo-woo way:
Being a "cut" itself is entirely equivalent to being the least upper bound of a cut. Every rational number, $q$ is equivalent to the cut $A_q = (-\infty, q)$ where $\sup A_q = q$. And as a collection the rational numbers with field operations and order, is completely equivalent to all such "rational cuts" as a collection with field operations with order.
And every real number $x$ is equivalent to the cut $A_x = (-\infty, ????)$ where $\sup A_x = x$. [In actuality, after we have defined the real numbers, we can see that $A_x = (-\infty, x)$; we jst had no way to express such an idea before we had the real numbers defined.] And as a collection, the real of real numbers (with field operations and order) is completely equivalent to the ordered field of all dedekind cuts.
So we think of the cuts and the least upper bounds of the cuts as being equivalent. In that way the $\mathbb Q$ (whether we think of it as a set of cuts or as the set of rational numbers that are least upper bounds of cuts) will be a subset of $\mathbb R$ (whether we think of it as a set of cuts or as the set or rational and/or irrational number that are the least upper bounds of cuts).
But of course if we think of $\mathbb Q$ is one way, and $\mathbb R$ in a different way, we get ... confusion.
There exists an injective relation from $\mathbb Q$ to $\mathbb R$
$$q\mapsto \{x \in \mathbb Q| x< q\}$$
and typically, we speak about the image of $\mathbb Q$ under that injection, not about $\mathbb Q$ itself, so we say that $\mathbb Q\subseteq\mathbb R$.