Find the supremum $(0,1)\setminus \mathbb{Q}$. My intuition tells me that $ A = (0,1)\setminus\mathbb{Q}$  has no supremum. But I'm not sure if the following arguments are sufficient or if my intuition is misguided...
We have that 1 is an upper bound of A, since:$$1 > a  \quad \forall a \in A$$
then because $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a $q$ in $\mathbb{Q}$ such that: 
$$a < q < 1 \quad \forall a \in A$$
hence we can find an $r$ in $\mathbb{Q}$ such that 
$$a < r < q \quad \forall a \in A$$
and the process repeats indefinitely, hence we will never find a least upper bound for the set. I assume I can use something similar to prove that $A$ has no infimum...
Any help is appreciated!

The set $A$ is non-empty and bounded above by $1$. It therefore has a supremum $\sup A \le 1$ in $\mathbb{R}$. You can also verify that 
$$1-\varepsilon < \sup A,$$ for all $\varepsilon > 0$. Therefore
$$
\sup A = 1.
$$
