How to construct a dense subset of $\mathbb R$ other than rationals.

First note that you can "cheat" by taking any subset and take a union with the rationals.

Second note that you can always cheat by taking some real number $x$ and considering the set $\{x+q\mid q\in\mathbb Q\}$. If $x$ is irrational then the set is not the rationals.

Now more seriously, you can note that the irrationals ($\mathbb R\setminus\mathbb Q$) are dense, as well all the irrational algebraic numbers ($\sqrt2$ and such). More interestingly the set $\{\sin n\mid n\in\mathbb N\}$ is dense in $[0,1]$ so it can be stretched (or multiplied) into a dense set of $\mathbb R$.

However an important fact is that every countable dense linear order is isomorphic to the rationals, so if your dense set is countable it will not differ too much from the rationals.

Let's construct dense sets in $[0,1]$. We can then get a dense set in $\Bbb R$ by taking the union of dense sets for the intervals $[n, n+1]$ (a dense set for $[n,n+1]$ can be obtained from a dense set of $[0,1]$ by shifting). To make things more interesting, we will find dense sets for which, given any two distinct elements in the set there is a number between them that is not in the set.

For a dense set in $[0,1]$, you can take:

The irrationals in $[0,1]$.

Or: take $[0,1]$. Take its midpoint $1/2$ to be an element in the, to be constructed, dense set. Then take the midpoints of $(0,1/2)$ and $(1/2,1)$ to be elements in the sense set. Then take the midpoints of the four sets obtained by splitting the two prior sets in two...

Or: use any similar, carefully done, construction similar to the preceding example. For instance, you could successively split $[0,1]$ as above (always splitting the previous sets in half), but choose irrationals in each piece.

Take any irrational number $\alpha$ and consider the set $E = \{n\alpha \bmod 1\ : n \in \mathbb{N}\}$. By the equidistribution theorem this set is uniformly distributed (and thus must be dense) on $[0,1]$. For a set dense on all of $\mathbb{R}$ take $\cup_{n \in \mathbb{Z}} (n + E)$.