Non-trivial open dense subset of $\mathbb{R}$.

$A = \mathbb{R} \setminus \{0\}$ works for this purpose, and isn't equal to $\mathbb{R}$. But that's still fairly trivial.

For a less trivial example, fix an enumeration $\{r_n\}_{n = 0}^{\infty}$ of rational numbers and a positive number $\epsilon$. Define open intervals

$$\mathcal{O}_n = \left(r_n - \frac{\epsilon}{2^{n + 2}}, r_n + \frac{\epsilon}{2^{n + 2}}\right)$$

and define $\mathcal{O} = \bigcup_n \mathcal{O}_n$. Then $\mathcal{O}$ is an open, dense subset of $\mathbb{R}$ with Lebesgue measure at most $\epsilon$.

In fact, we could (by dilating one of our intervals) make the measure of $\mathcal{O}$ equal to any given positive number $\epsilon$.

An important non-trivial example is the middle-thirds Cantor set $C$: it’s a subset of $\Bbb R$ of cardinality $2^\omega=\mathfrak{c}=|\Bbb R|$, and since it is closed and nowhere dense in $\Bbb R$, its complement is open and dense. With a fairly small change in the construction one can construct a ‘fat’ Cantor set that has all of the properties of $C$ that I just mentioned and in addition has positive Lebesgue measure; its complement is again a dense open subset of $\Bbb R$.

Take $A=\mathbb{R}-\{0\}$. This clearly open and dense.