We can use the Cantor-intersection theorem and the Archimedean axiom to prove the Dedekind's theorem (source:Enrico Gregorio's answer).
Suppose $A,B\subseteq\Bbb R$ are non-empty s. t. $(\forall a\in A)(\forall b\in B)\quad a\le B$ and $A\cup B=\Bbb R$. Then $\exists ! c\in\Bbb R$ s. t. $a\le c\le b, \forall a\in A,\forall b\in B$.
Let's abbreviate the above with $A\le B$ and $A\le c\le B$.
Since $A\ne\emptyset\space\land\space B\ne\emptyset$, let's choose $a_0\in A$ and $b_0\in B$.
Now, we choose $c_0=\frac{a_0+b_0}2$.
If $A\le c_0\le B$, we're done. Otherwise, it is either
$$c_0<a,\text{for some }a\in A\space\text {or}\space c_0>b\text{ for some }b\in B.$$
If $c_0<a$, then let $a_1=a, b_1=b_0$.
If $c_0>b$, then let $a_1=a_0, b_1=b$.
In either case, $a_0\le a_1\space\land\space b_0\ge b_1\implies [a_0,b_0]\supseteq[a_1,b_1]$.
Let $d=b_0-a_0$. We have $b_0-c_0=c_0-a_0=\frac{b_0-a_0}2=\frac{d}2$.
If $c_0<a$, then $\frac{d}2=b_0-c_0>b_0-a=b_1-a_1$.
If $c_0>b$, then $\frac{d}2=c_0-a_0>b-a_0=b_1-a_1$.
We can repeat the procedure to build a sequence of nested closed intervals:
$$[a_0,b_0]\supseteq[a_1,b_1]\supseteq\cdots\supseteq[a_n,b_n]$$
If at some point $A\le c_n\le B$, we're done. Otherwise, we get a sequence of closed nested intervals $[a_n,b_n]$ with $a_n\in A,b_n\in B$ and $b_n-a_n\le\frac{d}{2^n}.$
By the Cantor-intersection theorem (now assumed as an axiom), there is $c\in\bigcap_n[a_n,b_n]$.
Now we want to prove $A\le c\le B$. Let's assume the opposite: there is either
$$a\in A\text{ with } c<a\space\text{ or }\space b\in B\text{ with } c>b.$$
Suppose $a>c$.
Let's take $\varepsilon=\frac{a-c}2$.
We can take $n\in\Bbb N$ s. t. $\frac{d}{2^n}<\varepsilon$. $(*)$
Then we obtain:
$b_n<a+\varepsilon\le c+\varepsilon=c+\frac{a-c}2=\frac{a+c}2\le a$ which contradicts $A\le B$.
Analogously in case of $c>b$.
$(*)$ As $2^n>n, n\in\Bbb N$, we have $\frac1{2^n}<\frac1n\implies\frac{d}{2^n}<\frac{d}n$. In order to make $\frac{d}{2^n}<\varepsilon$, we could take $n$ s. t. $\frac{d}n<\varepsilon$, which is guaranteed to hold true by the Archimedean axiom.
Now, we want to prove the Dedekind's theorem is equivalent to the axiom of completeness (like Henno Brandsma did here):
Every non-empty subset of $\Bbb R$ that is bounded above has a supremum in $\Bbb R$.
Direction $\boxed{\Rightarrow}$
Suppose Dedekind's theorem holds.
Let $S\subset\Bbb R$ with some upper bound $u$ and let's define:
$$A=\{x\in\Bbb R\mid\exists s\in S, x<s\}\\\text{and}\\ B=\{x\in\Bbb R\mid\forall s\in S, x\ge s\}$$
$B\ne\emptyset$ because $u\in B$.
Then our $c=\sup S$.
Direction $\boxed{\Leftarrow}$
Assume the axiom of completeness holds.
Then we take $c=\sup A$.
Best Answer
The only difference between $\mathbb Q$ and $\mathbb R$ is the completeness axiom. So any property which is true in $\mathbb R$ but not in $\mathbb Q$ needs the completeness axiom. If you proof for example the existence of $\sqrt 2$ you will need the completeness axiom somewhere in your proof (otherwise you could prove $\sqrt 2 \in \mathbb Q$).
Now you can think of all theorems like the intermediate value theorem which hold for $\mathbb R$ as your basic number system but not for $\mathbb Q$. You will not have these theorems if you do not have the completeness axiom...
EDIT: Please have a look at these threads for your additional question: