The big difference between $\Bbb Q$ and $\Bbb R$ stays only in one simple property.
Without considering any particular construction, we can define $\Bbb Q$ to be the $\bf field$ $\bf of$ $\bf fraction$ $\bf of$ $\bf the$ $\bf integers$ $\Bbb Z$: I say 'the', because all fields of fractions are equal up to isomorphisms. So our $\Bbb Q$ happens to be an (infinite) ordered field with the Archimedean property;
Now, leave apart the Archimedean property, and consider an ordered field with the following additional property: $$\text{every non-empty bounded above set has a least upper bound.}$$
You've so just obtained $\Bbb R$! In fact, it can be shown that there indeed exists one such field. And all $\bf complete$ (with complete we refer to this last property) $\bf ordered$ $\bf field$ are equal up to isomorphism. The Archimedean property now follows automatically from the completeness. We can then think of $\Bbb Q$ to be a subset of $\Bbb R$, since $\Bbb Q$ is isomorphic to a subset of $\Bbb R$.
Archimedean/Archimedian property:
The Archimedean (or Archimedian) property basically says that the set $\Bbb N$ is not bounded above in $\Bbb R$. Surely, if we consider $\Bbb N$ by itself (i.e., not embedded in $\Bbb R$) as an ordered set, then it is unbounded above by definition (in fact, for each natural $n\in\Bbb N$, there exists also the number $n+1 \in\Bbb N$, which is defined to be greater than $n$).
Now, let's embed $\Bbb N$ in our $\Bbb R$. It can be done by induction. Let $\phi:\Bbb N\to \Bbb R$ be a map such that
\begin{equation}
\begin{cases}
\phi(1)=1\qquad\qquad\qquad\text{to 1 we associate the multiplicative identity element of the field $\Bbb R$};\\
\phi(n+1)=\phi(n)+1\qquad\text{the number 2 goes in $1+1$, the number 3 goes in $1+1+1$, $\dots$)}
\end{cases}
\end{equation}
(since now then we simply denote with $n$ the number $\phi(n)$ and we consider $\Bbb N$ only as a special subset of $\Bbb R$.)
As before, we know that there is no natural $m$ such that $m>=n$ for all $n\in \Bbb N$, in fact nothing has changed in the immersion of $\Bbb N$ into $\Bbb R$. But, how do we know that there isn't any $r\in \Bbb R$ such that $r>=n$ for all $n\in \Bbb N$? In other word, how can we be sure the set $\Bbb N$ in unbounded above in $\Bbb R$? Well, this is simply a consequence of the completeness property we assumed for $\Bbb R$.
*proof:*$\,\,\,\,\,$suppose $\Bbb N$ were bounded above. Since $\Bbb N\ne \emptyset$, there would be a least upper bound $\alpha$ for $\Bbb N$. Then
$$\alpha\ge n\qquad\text{for all $n$ in $\Bbb N$}.$$
Consequently
$$\alpha\ge n+1\qquad\text{for all $n$ in $\Bbb N$},$$ since $n+1$ is in $\Bbb N$ if $n$ is in $\Bbb N$. But this mean that $\alpha -1 \ge n$ for all $n$ in $\Bbb N$, and this mean that $\alpha -1$ is also an upper bound for $\Bbb N$, contradicting the fact that $\alpha$ is the least upper bound. $\blacksquare$
One can show that this property is equivalent to the fact that, given an $\varepsilon>0$ it exist a natural $n$ such that $\dfrac{1}{n}<\varepsilon$.