You can't really show that $\mathbb R$ is uncountable using Dedekind cuts. If you can show that, then I am unaware of any such proof (and having a lot of work around the intro to set theory course in my university for the past few years, I doubt that I would not have known such proof). You can show that $\mathcal P(\mathbb N)$ is uncountable by this method though.
Dedekind cuts provide a method for constructing and defining the real numbers, rather than a direct method of proving their size is $2^{\aleph_0}$.
However in proving that $|\mathbb R|=2^{\aleph_0}=|\mathcal P(\mathbb N)|$ we can use Dedekind cuts to establish the $\leq$ direction by showing that if we fix an enumeration of $\mathbb Q$ as $q_n$, then the sets $A_r = \{n\in\mathbb N\mid q_n<r\}$ for $r\in\mathbb R$ form an injection from $\mathbb R$ into $\mathcal P(\mathbb N)$. Therefore there are at most $2^{\aleph_0}$ many real numbers.
The other direction usually involves the Cantor set construction and shows that there are exactly $2^{\aleph_0}$ many real numbers.
I’ll show that the sum of two rational cuts is rational. Suppose that $A$ and $B$ are rational cuts; then there are $a,b\in\Bbb Q$ such that $A=\{q\in\Bbb Q:q<a\}$ and $B=\{q\in\Bbb Q:q<b\}$. By definition $$A+B=\{p+q:p\in A\text{ and }q\in B\}\;;$$ I’ll prove that $A+B=\{q\in\Bbb Q:q<a+b\}$, the Dedekind cut corresponding to the rational number $a+b$.
Suppose that $r\in A+B$. Then there are $p\in A$ and $q\in B$ such that $r=p+q$. Since $p\in A$, we know that $p<a$, and similarly, since $q\in B$, we know that $q<b$, so $r=p+q<a+b$. This shows that $A+B\subseteq\{q\in\Bbb Q:q<a+b\}$.
Now suppose that $r\in\Bbb Q$ and $r<a+b$. Let $d=\frac12(a+b-r)>0$, and note that $d$ is rational. Let $p=a-d$ and $q=b-d$; then $p,q\in\Bbb Q$. Moreover $p<a$ (since $d>0$), so $p\in A$, and $q<b$, so $q\in B$, and $$p+q=(a-d)+(b-d)=a+b-2d=a+b-(a+b-r)=r\;.$$ Thus, $r\in A+B$, and we’ve shown that $\{q\in\Bbb q:q<a+b\}\subseteq A+B$. Putting the pieces together, we conclude that $A+B=\{q\in\Bbb Q:q<a+b\}$. $\dashv$
Here I didn’t have to prove directly that $A+B$ was a Dedekind cut, because I could show that it was equal to something known to be a Dedekind cut. You should try to prove that for all Dedekind cuts $A$ and $B$, $A+B=\{p+q:p\in A\text{ and }q\in B\}$ is a Dedekind cut.
- It’s easy to show that $A+B\ne\varnothing$.
- Suppose that $r\in A+B$, $s\in\Bbb Q$, and $s\le r$; you need to show that $s\in A+B$. You know that $r=p+q$ for some $p\in A$ and $q\in B$. Let $d=\frac12(r-s)$, and consider the rational numbers $p-d$ and $q-d$.
- Suppose that $r\in A+B$; you need to show that there is an $s\in A+B$ such that $r<s$. Start by writing $r=p+q$ for some $p\in A$ and $q\in B$, and use the fact that $A$ and $B$ have no largest elements.
Once you’ve done that, you can show that if $A$ is a rational cut and $B$ is an irrational cut, then $A+B$ is irrational. You already know that $A+B$ is a Dedekind cut, so you just have to show that it’s not a rational Dedekind cut: for each $r\in\Bbb Q$, $A+B\ne\{q\in\Bbb Q:q<r\}$. This can be done by contradiction: show that if $A+B=\{q\in\Bbb Q:q<r\}$ for some $r\in\Bbb Q$, then the cut $B$ is rational.
When both $A$ and $B$ are irrational cuts, $A+B$ can be either rational or irrational, depending on exactly which cuts $A$ and $B$ are; you can’t prove any general conclusion here.
Best Answer
When you say "lesser than the resulting real number", you think as if that real number (i.e. the sum of the two given reals) was already known/obtained by other means.
This is not the case, it is only known from the definition, which you don't have to prove.
To illustrate, $\sqrt2+\sqrt3$ is the set of rational numbers $p+q$ such that $p^2\le2$ and $q^2\le3$, and nothing else. And obviously, no larger rational belongs to that set by definition.
As long as you have not defined the Dedekind cuts, there are no reals, and as long as you have not defined the addition of Dedekind cuts, the sum of two reals doesn't exist.
A philosophical remark:
You might be tempted to say, yeah, but we know that the reals are there independently of us, and $\sqrt2+\sqrt3$ exists and could be such that Dedekind addition fails.
That's right, there could exist some real number $\sqrt2+\sqrt3$ in some real number theory, such that there is a rational that invalidates the inequality. But we need to use definitions to specify which number theory we want to deal with, namely one where Dedekind addition is foolproof (if that exists).